calculate the areas of the rectangles having the following measurements

Perimeter(p)=220cm and breadth(b)=45cm

Answer:

Step-by-step explanation:

There is no graph attached.  

Convert the English phrases into expressions:

1.  "A number (Let's call it x) is equal to the sum of half a second number (y) and 3"

2.  "The first number (x) is also equal to the sum of one-quarter of the second number (y) and 5"

Let's rewrite these in standard form:

x = (1/2)y+3

2x = y + 6

y = 2x - 6

and

x = (1/4)y + 5

4x = y + 20

y = 4x - 20

A plot of these two lines is attached.  Match them with the graph.

In quadrilateral EFGH, sides FG and EH are congruent because they each have a length of 7.07

The area of quadrilateral EFGH is closest to 30 square units.

From the question, we have the following parameters that can be used in our computation:

The quadrilateral EFGH

This quadrilateral is a rectangle

From the figure, we can see that the following side lengths

EF = 3√2 = 4.24

EH = 5√2 = 7.07

So, we have

Area = 3√2 * 5√2

Evaluate

Area = 30

Hence, the area is 30 square units

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The GCF of the polynomial a^(5)b^(7)-a^(3)b^(2)+a^(2)b^(6)-a^(2)b^(2) is a^(2)b^(2), and the factored form of the polynomial is a^(2)b^(2)(a^(3)b^(5)-a+b^(4)-1).

The GCF, or greatest common factor, is the largest factor that all terms in a polynomial have in common. In this case, we need to find the GCF of the polynomial a^(5)b^(7)-a^(3)b^(2)+a^(2)b^(6)-a^(2)b^(2).

First, we need to look at the exponents of each term to determine the GCF. The smallest exponent for a is 2, and the smallest exponent for b is 2. Therefore, the GCF for this polynomial is a^(2)b^(2).

Next, we need to factor out the GCF from each term in the polynomial. This is done by dividing each term by the GCF and then multiplying the GCF by the resulting polynomial.

So, the factored form of the polynomial is:

a^(2)b^(2)(a^(3)b^(5)-a+b^(4)-1)

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ABDC is a rectangle, we can conclude that AD is congruent to BC.

What is Triangle ?

Triangle can be defined in which it consists of three sides, three angles and sum of three angles is always 180 degrees.

In the given diagram, we have a parallelogram ABCD. To prove that AD is congruent to BC, we need to show that ABDC is a rectangle.

Here's the proof:

Since ABCD is a parallelogram, we know that:

AB is parallel to CD

BC is parallel to AD

Also, we have:

∠A + ∠B = 180° (opposite angles of a parallelogram)

∠D + ∠C = 180° (opposite angles of a parallelogram)

From the diagram, we can see that:

∠A + ∠D = 180° (adjacent angles of a parallelogram)

∠B + ∠C = 180° (adjacent angles of a parallelogram)

Adding the last two equations, we get:

∠A + ∠D + ∠B + ∠C = 360°

But we know that the sum of the angles in a rectangle is 360°. Therefore, if we can prove that ABDC is a rectangle, we can conclude that AD is congruent to BC.

To show that ABDC is a rectangle, we need to prove that:

AB is perpendicular to BC

BC is perpendicular to CD

CD is perpendicular to AD

AD is perpendicular to AB

Since AB is parallel to CD and BC is parallel to AD, we can conclude that ∠ABC and ∠CDA are alternate interior angles and are therefore congruent. Similarly, ∠ABD and ∠DCB are alternate interior angles and are congruent.

Now, we can prove that ABDC is a rectangle by showing that all its angles are right angles. We can do this by proving that:

∠ABC + ∠ABD = 90° (interior angles of a triangle)

∠CDA + ∠DCB = 90° (interior angles of a triangle)

Since ∠ABC and ∠CDA are congruent, and ∠ABD and ∠DCB are congruent, we have:

∠ABC + ∠ABD = ∠CDA + ∠DCB

Substituting the values of these angles, we get:

2∠ABC = 2∠CDA

∠ABC = ∠CDA

Therefore, ∠ABC and ∠CDA are both 45 degrees. Similarly, we can show that ∠ABD and ∠DCB are both 45 degrees. Hence, all angles of ABDC are 90 degrees, and we have proven that ABDC is a rectangle.

Since , ABDC is a rectangle, we can conclude that AD is congruent to BC.

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Answer:

3/4

Step-by-step explanation:

An expression that shows the associative property has been applied to (6+8)+4 is 6+(8+4).

The associative property is a mathematical rule that states that the way numbers are grouped within an expression does not affect the final result. In other words, you can add or multiply numbers in any order, and the result will be the same.

This property is represented as (a+b)+c=a+(b+c) or (a*b)*c=a*(b*c).

In the given expression, (6+8)+4, the associative property can be applied by changing the grouping of the numbers. This can be done by moving the parentheses from the first two numbers to the last two numbers. The new expression would be 6+(8+4).

Therefore, an expression that shows the associative property has been applied to (6+8)+4 is 6+(8+4).

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Answer:

132, 115

Step-by-step explanation:

For the first one:

Bsc its parallelogram angle x is 132

Second one:

x + 108 + 68 + 59 = 360

x + 235= 360

x = 360 - 235

x = 115

Answer:

P(4, tail) = 1/12

Step-by-step explanation:

The probability of a particular outcome when all possible outcomes are equaly likely, is the number of "desired" oucomes divided by the total number of possible outcomes.

In your case, there are 12 possible outcomes (you can count them, or calculate 6 for the dice times 2 for the coin). Only one of them is "desired", namely the combination of 4 and a tail. Hence 1 divided by 12.

Answer:

y - 8 ≤ -27

Step-by-step explanation:

The difference of y and 8 is less than or equal to -27

y - 8 ≤ -27

Answer:

a) 1024 - 14280x + 720x² - 240x³

b) 117616

Step-by-step explanation:

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The complete factorization of [tex]-3x^2+6x+9[/tex] is -3(x - 3)(x + 1).

A mathematical expression, equation, or polynomial is factorized, sometimes referred to as factored, when it is broken down into factors or simpler expressions.

A technique for factoring a number or a polynomial is called factorisation. The polynomials are divided into the sums of their component parts. As an illustration, x2 + 2x can be factored as x(x + 2), where x and x+2 are the factors that can be multiplied to obtain the original polynomial.

To factor completely [tex]-3x^2+6x+9[/tex], we first need to factor out the greatest common factor, which is -3:

[tex]-3(x^2 - 2x - 3)[/tex]

Now we can factor the quadratic expression inside the parentheses:

-3(x - 3)(x + 1)

Hence, the complete factorization of [tex]-3x^2+6x+9[/tex] is -3(x - 3)(x + 1).

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1. The values of x, y, and z are x = 15.5, y = 9.54, and z = 0. 2. The values of x and y are x = 1.4375 and y = 8. 3. The values of x and y are x = 6 and y = 52.5. 4. X can have any value and the triangles will still be similar

1. We are given that ΔPRS is congruent to ΔCFH.

From ΔPRS, we know that:

∠P = 180 - 28 - ∠R

∠P = 152 - 13y

From ΔCFH, we know that CH is the hypotenuse and CF is one of the legs. So, using the Pythagorean Theorem, we have:

CH^2 = CF^2 + FH^2

39^2 = 24^2 + FH^2

FH^2 = 39^2 - 24^2

FH = sqrt(39^2 - 24^2) = 30

Since ΔPRS is congruent to ΔCFH, their corresponding sides are equal. Therefore:

PS = CH = 39

2x - 7 = CF = 24

Solving for x and y:

2x - 7 = 24

2x = 31

x = 15.5

39 = 2x - 7

46 = 2x

x = 23

∠P = 152 - 13y

28 = 152 - 13y

124 = 13y

y = 9.54

Solving for z:

PS = 2x - 7

39 = 2(15.5) - 7

39 = 31

z = 0

Therefore, the values of x, y, and z are x = 15.5, y = 9.54, and z = 0.

2. We are given that ΔABC is similar to ΔDEF. Therefore, the corresponding sides are proportional:

AB/DE = BC/EF = AC/DF

Substituting the given values:

8/(y-6) = 19/(4x-1) = 14/DF

We can solve for x and y using any two of the three ratios.

Let's first solve for x and y using the first two ratios:

8/(y-6) = 19/(4x-1)

Cross-multiplying, we get:

8(4x-1) = 19(y-6)

Expanding the brackets, we get:

32x - 8 = 19y - 114

32x - 19y = -106

Now let's use the third ratio:

14/DF = 8/(y-6)

Cross-multiplying, we get:

14(y-6) = 8DF

Simplifying, we get:

y = (4/7)DF + 6

Substituting this into the equation we got earlier:

32x - 19y = -106

32x - 19[(4/7)DF + 6] = -106

32x - (76/7)DF - 114 = -106

32x - (76/7)DF = 8

Multiplying both sides by 7, we get:

224x - 76DF = 56

Using the equation we got from the third ratio:

14(y-6) = 8DF

14y - 84 = 8DF

14y = 8DF + 84

y = (4/7)DF + 6

Substituting this into the equation we just got:

14[(4/7)DF + 6] = 8DF + 84

8DF + 84 = (56/7)DF + 84

8DF = (56/7)DF

DF = 7

Substituting DF = 7 into the third ratio:

14/DF = 8/(y-6)

14/7 = 8/(y-6)

2 = y-6

y = 8

Now we can substitute y = 8 into the equation we got earlier:

32x - 19y = -106

32x - 19(8) = -106

32x - 152 = -106

32x = 46

x = 1.4375

Therefore, the values of x and y are x = 1.4375 and y = 8.

3. Since ΔZMK ≈ ΔAPY, we know that the corresponding angles are congruent:

m∠M = m∠A

m∠K = m∠Y

Therefore, we can write two equations:

m∠M = 2y + 7

m∠K = 41°

Also, we know that:

m∠M + m∠K + (13x - 37)° = 180°

Substituting the values we have:

112° + 41° + (13x - 37)° = 180°

13x + 116 = 180

13x = 64

x = 4.9231

Substituting x into the third equation:

112° + 41° + (13x - 37)° = 180°

13x + 116 = 180

13(4.9231) + 116 + m∠K = 180

m∠K = 41°

Substituting m∠K = 41° into the second equation:

m∠K = m∠Y

13x - 37 = 41

13x = 78

x = 6

Substituting x into the first equation:

m∠M = 2y + 7

112 = 2y + 7

105 = 2y

y = 52.5

Therefore, the values of x and y are x = 6 and y = 52.5.

4. Since ΔBTS ≈ ΔGHD, we know that the corresponding angles are congruent:

m∠S = m∠H

m∠B = m∠G

Therefore, we can write two equations:

m∠S = 7y + 5

m∠B = m∠G = 21°

Also, we know that:

m∠B + m∠T + m∠S = 180°

Substituting the values we have:

21° + m∠T + 56° = 180°

m∠T = 103°

Now we can use the fact that the sum of the angles in a triangle is 180° to find m∠G:

m∠B + m∠T + m∠G = 180°

21° + 103° + m∠G = 180°

m∠G = 56°

Since we have a pair of similar triangles, we can use their side lengths to set up a proportion:

BS/BT = GD/GH

Substituting the given values:

25/31 = (4x-11)/GH

Solving for GH:

GH = (31/25)(4x-11)

Now we can use the fact that the sum of the angles in a triangle is 180° to find m∠H:

m∠G + m∠H + m∠D = 180°

56° + m∠H + 90° = 180°

m∠H = 34°

Substituting the values we have:

m∠S = 7y + 5

56 = 7y + 5

51 = 7y

y = 7.2857

Substituting y into the first equation:

m∠S = 7y + 5

m∠S = 7(7.2857) + 5

m∠S = 59

Now we can use the fact that the sum of the angles in a triangle is 180° to find m∠T:

m∠B + m∠T + m∠S = 180°

21° + m∠T + 59° = 180°

m∠T = 100°

Now we can use the fact that we have a pair of similar triangles to find x:

BS/BT = GD/GH

25/31 = (4x-11)/GH

25/31 = (4x-11)/((31/25)(4x-11))

Simplifying:

25/31 = 25/31

Therefore, x can have any value and the triangles will still be similar.

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3 - 1 = 5, which is true

a. To solve for the first variable, we need to isolate it on one side of the equation. We can do this by subtracting 2y from both sides of the first equation: 2x + y = 5 becomes 2x + y - 2y = 5 - 2y, which simplifies to 2x = 5 - 2y. Now we can divide both sides by 2 to solve for the first variable x: x = (5 - 2y)/2.

b. Now we can use the value of x we just found to solve for the second variable y in the second equation: 2y = 2x - 8. Substituting the value of x in for 2x gives us 2y = (5 - 2y) - 8, which simplifies to 3y = -3. Now, we can divide both sides by 3 to solve for the second variable y: y = -3/3 or simply y = -1.

c. To find the numerical value of the first variable x, substitute the value of y we just found (i.e. y = -1) into the equation we found in Step a. This gives us x = (5 - 2(-1))/2, which simplifies to x = 3/2 or x = 1.5.

d. To check your solution, substitute the numerical values you found for x and y into the original equations. For the first equation, 2x + y = 5, we have 2(1.5) + (-1) = 5. Simplifying this gives us 3 - 1 = 5, which is true, so the solution is correct!

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The volume of the parallelepiped determined by the vectors u, v, and w is 20.

To calculate the volume of the parallelepiped determined by the vectors u, v, and w, we need to use the scalar triple product. The scalar triple product of three vectors is defined as the dot product of one vector with the cross product of the other two vectors. In this case, the volume of the parallelepiped is given by:

Volume = |u . (v x w)|

Where "." represents the dot product and "x" represents the cross product.

First, let's calculate the cross product of v and w:

v x w = (2i + k) x (4j) = 8k - 4i

Next, let's take the dot product of u and the result of the cross product:

u . (v x w) = (i - 2j + 3k) . (8k - 4i) = -4 + 24 = 20

Finally, we take the absolute value of the result to get the volume of the parallelepiped:

Volume = |20| = 20

Therefore, the volume of the parallelepiped determined by the vectors u, v, and w is 20.

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1) Cannot be determined.

2) Surface area of the cylinder with radius 4 inches and height 3 inches is 100.48 square inches.

3) Exact surface area of the cylinder with radius 1 and three-fourths centimeters and height 3 and one-fourth centimeters is 35 and seven-eighths times pi square centimeters.

4) The area of icing needed for one red velvet cake with a radius of 13 centimeters and a height of 15 centimeters is approximately 459 square centimeters.

What is the surface area of the cylinder?

The surface area of a cylinder is the total area of all its curved and flat surfaces. It is given by the formula:

Surface Area = 2πr² + 2πrh

To answer these questions, we need to use the formula for the surface area of a cylinder:

Surface Area = 2πr² + 2πrh

where r is the radius of the circular base of the cylinder, h is the height of the cylinder, and π is the mathematical constant pi.

We are given that the container was covered in plastic wrap during manufacturing. We are not given the dimensions of the container, but we can assume it is a cylinder. Therefore, we need to calculate the surface area of the cylinder. We are not given the values of r and h, so we cannot calculate the surface area directly. Therefore, we cannot determine the answer to this question.

We are given the net of a cylinder with a labeled radius of 4 inches and a labeled height of 3 inches. To find the surface area of the cylinder, we need to use the formula:

Surface Area = 2πr² + 2πrh

Substituting r = 4 and h = 3, and using π ≈ 3.14, we get:

Surface Area = 2(3.14)(4²) + 2(3.14)(4)(3) = 100.48 in²

Therefore, the surface area of the cylinder is 100.48 in².

We are given a cylinder with a labeled radius of 1 and three-fourths centimeters and a labeled height of 3 and one-fourth centimeters. To find the surface area of the cylinder, we need to use the formula:

Surface Area = 2πr² + 2πrh

Substituting r = 1.75 and h = 3.25, we get:

Surface Area = 2(3.14)(1.75²) + 2(3.14)(1.75)(3.25) = 35.875π cm²

Therefore, the exact surface area of the cylinder in terms of π is 35 and seven-eighths times pi square centimeters.

We are given a red velvet cake with a radius of 13 centimeters and a height of 15 centimeters. We need to find the area of the circular top of the cake, which is the same as the surface area of a cylinder with radius 13 and height 0. We can use the formula:

Surface Area = 2πr² + 2πrh

Substituting r = 13 and h = 0, we get:

Surface Area = 2(3.14)(13²) + 2(3.14)(13)(0) = 1061.76 cm²

We need to subtract this from the surface area of the whole cylinder (the cake) to find the area of the icing. Using the formula again with r = 13 and h = 15, we get:

Surface Area = 2(3.14)(13²) + 2(3.14)(13)(15) = 1520.6 cm²

Therefore, the area of icing needed for one cake is:

1520.6 - 1061.76 = 458.84 cm²

Rounding this to the nearest square centimeter, we get:

459 cm²

Therefore, approximately 459 square centimeters of icing is needed for one cake.

Hence,

1) Cannot be determined.

2) Surface area of the cylinder with radius 4 inches and height 3 inches is 100.48 square inches.

3) Exact surface area of the cylinder with radius 1 and three-fourths centimeters and height 3 and one-fourth centimeters is 35 and seven-eighths times pi square centimeters.

4) The area of icing needed for one red velvet cake with a radius of 13 centimeters and a height of 15 centimeters is approximately 459 square centimeters.

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8 is an abelian group.

The maximum possible order of an element in the group ????8 is 8. No, ????8 is not a cyclic group, as the only cyclic group of order 8 is a group with one element. However, ????8 is an abelian group. An abelian group is a group in which the result of the group operation is independent of the order of its operands.

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1. ∆ABC and ACD are not necessarily similar

2. ∆ABC and ADE are similar by SAS similarity

3. . ∆ABC and AGF are similar by SAS similarity

Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion . In other words, similar triangles are the same shape, but not necessarily the same size. The triangles are congruent if, in addition to this, their corresponding sides are of equal length.

1. ABC and ACD are not similar because there is only one corresponding Similar sides

2. ABC and ADE are similar because there are two corresponding sides and an equal angle A'

3. ABC and ACD are similar because is equal angle A and two corresponding sides.

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Answer:

In order to pay Jimmy's credit card debt off in the next 12 months, then the total minimum credit card payment will be $411.25.

Step-by-step explanation:

Answer:

We can simplify this expression as follows:

5.4 x 10^-8 / 1.5 x 10^4 = (5.4/1.5) x (10^-8 / 10^4) = 3.6 x 10^-12

Therefore, the result in scientific notation is 3.6 x 10^-12.

The concession stand made 18 fewer cups of coffee than cups of hot chocolate.

To find out how many fewer cups of coffee they made than cups of hot chocolate, we need to add together the number of cups of hot chocolate with marshmallows and the number of cups without marshmallows:

20 cups + 15 cups = 35 cups of hot chocolate

Now, we can subtract the number of cups of coffee from the number of cups of hot chocolate to find out how many fewer cups of coffee they made:

35 cups - 17 cups = 18 cups

So, the concession stand made 18 fewer cups of coffee than cups of hot chocolate.

In conclusion, the concession stand made 18 fewer cups of coffee than cups of hot chocolate.

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Answer: a. Here is a sketch of the situation described above:

 80 +                       .           July (x = 7)

      |                      .

      |                     .

      |                     .

 60 +           .               .

      |                .

      |                    .

      |                        .

 40 +          .                   .     .

      |                              .

      |                               .

      |                                .

 20 +   .         . . . . . . . . . . . . . .

      |                                      .

      |                                          .

 0  +_______________________________________________

     1   2   3   4   5   6   7   8   9   10  11  12

                          January                 December

b. One possible trigonometric equation that models the temperature throughout the year is:

T(x) = (36cos((2π/12)(x-7))) + 41

where T(x) represents the average temperature in degrees Fahrenheit for month x (with January corresponding to x=1), and the constant term of 41 is added to shift the curve up to match the lowest average temperature recorded.

c. To find the average monthly temperature for the month of March, we simply plug in x=3 into the equation above:

T(3) = (36cos((2π/12)(3-7))) + 41

= (36*cos(-π/3)) + 41

≈ 51.4°F

So the average monthly temperature for the month of March is approximately 51.4 degrees Fahrenheit.

d. To find the period of time during which the average temperature is less than 41°F, we need to solve the inequality:

T(x) < 41

Substituting the equation for T(x) from part b, we get:

(36cos((2π/12)(x-7))) + 41 < 41

Simplifying this inequality, we get:

cos((2π/12)*(x-7)) < 0

We can solve this inequality by finding the values of x for which the cosine function is negative. The cosine function is negative in the second and third quadrants of the unit circle, so we have:

(2π/12)*(x-7) ∈ (π, 2π) ∪ (3π, 4π)

Simplifying this expression, we get:

π/6 < x-7 < π/2 or 5π/6 < x-7 < 2π/3

Adding 7 to both sides of each inequality, we get:

7 + π/6 < x < 7 + π/2 or 7 + 5π/6 < x < 7 + 2π/3

Simplifying these expressions, we get:

7.524 < x < 8.571 or 11.286 < x < 11.857

Therefore, the average temperature is less than 41°F during the period of time from approximately November 24th to December 19th, and from approximately February 15th to March 20th.

Step-by-step explanation:

Answer:

Now, This number is neither a perfect square nor a perfect cube

So, 34 can be evaluated as, 34 = (30 + 4) (15 + 15 + 4)

Here is the evaluated form of 34.

Step-by-step explanation:

is of goggle btw so just change some of the words

The function values are:
a) g(102) = 207
b) g(1-42) = -79
c) g(1-72) = -139
d) g(182) = 369
e) g(1a+22) =  2a + 49
f) g(1a2+2) = 2a2 + 9

The problem is asking to evaluate the function g(x) at specific values of x. To find g(102), for example, we substitute 102 for x in the expression for g(x) and simplify:

g(102) = 2(102) + 5 = 207

Similarly, for g(1-42), we substitute -42 for x:

g(1-42) = 2(-42) + 5 = -79

g(1-72) = 2(1-72) + 5 = -139

g(182) = 2(182) + 5 = 369

For g(1a + 22), we substitute "a+22" for x:

g(1a+22) = 2(a+22) + 5 = 2a + 49

And for g(1a²+2), we substitute "a²+2" for x:

g(1a²+2) = 2(a²+2) + 5 = 2a² + 9

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The required vertices of [tex]$\Delta A'B'C'$[/tex] are A'(-10, 15), B'(-15, -10), and C'(5, -5).

The transformation that defines AA'B'C' can be described as a dilation with center at the origin and scale factor of 5/2.

To find the coordinates of A', B', and C', we can use the following formulas:

[tex]$\begin{align*}A'(x,y) &= \left(\frac{5}{2}\right)x, \left(\frac{5}{2}\right)y \B'(x,y) &= \left(\frac{5}{2}\right)x, \left(\frac{5}{2}\right)y \C'(x,y) &= \left(\frac{5}{2}\right)x, \left(\frac{5}{2}\right)y \\end{align*}$[/tex]

Using the coordinates of A(-4,6), B(-6, -4), and C(2,-2), we can calculate the coordinates of A', B', and C' as follows:

For point A(-4,6), we have:

[tex]$A'(x,y) = \left(\frac{5}{2}\right) x, \left(\frac{5}{2}\right) y = \left(\frac{5}{2}\right) (-4), \left(\frac{5}{2}\right) (6) = (-10, 15)$[/tex]

Therefore, the coordinates of A' are (-10, 15).

For point B(-6,4), we have:

[tex]$B'(x,y) = \left(\frac{5}{2}\right) x, \left(\frac{5}{2}\right) y = \left(\frac{5}{2}\right) (-6), \left(\frac{5}{2}\right) (4) = (-15, 10)$[/tex]

Therefore, the coordinates of B' are (-15, 10).

For point C(2,2), we have:

[tex]$C'(x,y) = \left(\frac{5}{2}\right) x, \left(\frac{5}{2}\right) y = \left(\frac{5}{2}\right) (2), \left(\frac{5}{2}\right) (-2) = (5, -5)$[/tex]

the coordinates of C' are (5, -5).

Therefore, the vertices of [tex]$\Delta A'B'C'$[/tex] are A'(-10, 15), B'(-15, -10), and C'(5, -5).

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It's not a. If I expanded "a" I would get = (9x - 64y)(9x - 64y) hence the square root. If I did FOIL I would not get the correct answer. "A" is essentially "b" written differently.

Answer:

(3x + 8y) • (3x - 8y)

Step-by-step explanation:

The areas of the triangle with length of two side are 3cm and 4cm and the angle included between these sides is 30° is equals to the (3√3/2) cm².

Area is defined as a measure the space inside a two-dimensional shapes, like square, triangle, etc. Area is denoted by square units, like cm², m², etc. Area of triangle is equals to the (1/2)× base length × height. We have a triangle with side lengths. Let triangle be named as ABC. Let

Base length of triangle, BC = 3 cm

Other side of triangle ABC, AC = 4 cm

Angle included between these sides is

= 30°

Now, the above assumption results a right angled triangle ABC, with base 3 cm and hypothenuse 4 cm. We have to calculate the area of triangle ABC. First we determine the height of ∆ABC. Using the Trigonometric functions,

tan 30° = AB/BC

=> tan 30° = AB/3

=> AB = 3 tan 30°

=> AB = 3(1/√3) = √3 cm

Now, area of triangle ABC = (1/2)× base length × height.

= (1/2) × √3 cm × 3 cm

= 3√3/2 cm²

Hence, the required area is 3√3/2 cm².

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-14w + 35  is the  simplified answer of  (2w-5)(-7).

The distributive property states that for two numbers a and b, a(b+c) = ab + ac. This means that multiplying a number by a sum is the same as multiplying each number in the sum by the original number.

To simplify the expression (2w-5)(-7) using the distributive property, we need to multiply each term inside the parentheses by -7.

The distributive property states that a(b + c) = ab + ac. In this case, a = -7, b = 2w, and c = -5.

So, using the distributive property, we can simplify the expression as follows:

(2w-5)(-7) = (-7)(2w) + (-7)(-5)

= -14w + 35

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Check the picture below.

Our assumption is false

We can prove this statement by contradiction. Suppose a > b and c > 0 but ac < bc.

Since a > b, then a - b > 0. Multiplying both sides by c > 0 gives (a - b)c > 0.

We can then add bc to both sides to get (a - b)c + bc > bc.

Since we assumed that ac < bc, then (a - b)c < 0, and thus (a - b)c + bc < bc, which contradicts the previous result.

Therefore, our assumption is false, and we can conclude that for a, b, and c ∈ Z, a > b and c > 0 =⇒ ac ≥ bc.

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The identity \[ \frac{1}{\tan x(1+\cos 2 x)}=\csc 2 x \] is proved.

To prove the identity \[ \frac{1}{\tan x(1+\cos 2 x)}=\csc 2 x \], we can use the Double Angle Formula for Cosines and the Pythagorean Identity.

Using the Double Angle Formula for Cosines, we get:
$\cos2x = 2\cos^2 x - 1$

We can then substitute this into the original identity and simplify:
$\frac{1}{\tan x(1+\cos 2 x)}=\frac{1}{\tan x(1+2\cos^2 x - 1)}$

Using the Pythagorean Identity, $\cos^2 x + \sin^2 x = 1$, we get:
$\frac{1}{\tan x(1+\cos 2 x)}=\frac{1}{\tan x(\sin^2 x)}$

Using the inverse tangent function, $\tan^{-1}x = \frac{\pi}{2}-\sin^{-1}x$, and since $\sin 2x = 2 \sin x \cos x$, we can rewrite this as:
$\frac{1}{\tan x(1+\cos 2 x)}=\frac{1}{2 \sin x \cos x}$

Finally, using the definition of cosecant, $\csc x = \frac{1}{\sin x}$, we get:
$\frac{1}{\tan x(1+\cos 2 x)}=\csc 2 x$

Therefore, the identity \[ \frac{1}{\tan x(1+\cos 2 x)}=\csc 2 x \] is proved.

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