what is the ratio of 2:35

Answer:

Step-by-step explanation:

2b + 3d = 13.25

2b + 2d = 11.50

2b + 3d = 13.25

-2b - 2d = -11.50

d = $1.75 for one drink

2b + 3.50 = 11.50

2b = 8

b = $4 for one Big Mac

The coefficient of the first term is 80, and the degree is 9; the coefficient of the second term is -6, and the degree is 4; the coefficient of the third term is -8, and the degree is 0. The polynomial is 80x^(9)y^(2)-6x^(4)yz-8.

The coefficient of a term in a polynomial is the number that is multiplied by the variable(s) in the term. The degree of a term is the sum of the exponents of the variables in the term.

In the first term, 80x^(9)y^(2), the coefficient is 80 and the degree is 9 + 2 = 11.

In the second term, -6x^(4)yz, the coefficient is -6 and the degree is 4 + 1 + 1 = 6.

In the third term, -8, the coefficient is -8 and the degree is 0 since there are no variables.

The polynomial is already in standard form, so there is no need to find the polynomial. It is simply 80x^(9)y^(2)-6x^(4)yz-8.

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Answer:

Step-by-step explanation:

The llsab have to be divided by the number quotient and then once your sllab get hot then you answer.

READ THIS BACKWARDS

answer is adjacent angles

Answer:

complementary and

adjacent

Step-by-step explanation:

Complementary means the angles add up to 90°, that is, they make a right angle.

Adjacent means they are next to each other.

Both of these are the correct answer.

(since the question has boxes on the answers, you can mark more than one correct answer)

We can find g(x) for any value of x by using the function notation g(x) = f(x) + 6.

The function g is related to the parent function f by a vertical shift of 6 units. In function notation, we can write g in terms of f as:

g(x) = f(x) + 6

This means that for any value of x, we can find the corresponding value of g by first finding the value of f at that x value, and then adding 6.
For example, if x = 2, we can find g(2) by first finding f(2) and then adding 6:
f(2) = 2^(2) = 4
g(2) = f(2) + 6 = 4 + 6 = 10
So g(2) = 10.

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The principal amount lent out at simple interest is RS.690 in 3 years and RS.750 at the end of the second year RS.738.

To calculate the amount of money that needs to be lent out, use the following formula:

Amount = Principal * (1 + (Rate * Time))

Where:
Principal = 690
Rate = 0.05 (5% simple interest rate)
Time = 2 years

Therefore, the amount to be lent out is:
Amount = 690 * (1 + (0.05 * 2)) = 738

Therefore, the amount to be lent out is RS.738.

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Answer:

P'(1,2) S'(2,5)

Step-by-step explanation:

P(3,-2) S(4,-5) glode translation (x-2,y)

-2 -2

P(1,-2) S(2,-5) reflect over x-axis

P'(1,2) S'(2,5) ----> answer

hope this helpzzz

 a_(n) = -12 + 100n.

The recursive formula for the sequence is a_(n) = a_(n-1) + 100. The first term of the sequence is a_(1) = -12 and the common difference is 100. The explicit form for this sequence is a_(n) = -12 + 100n.

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The final answers are:

Example 4. 45: 1

46: 1

47: \( \frac{17}{12} \)

To evaluate each expression, we can use the identities for sine and cosine, and then simplify.

For example 4. 45, we have:

\( \sin ^{2} 120^{\circ}+\cos ^{2} 120^{\circ} \)

= \( (\frac{\sqrt{3}}{2})^{2}+(-\frac{1}{2})^{2} \)

= \( \frac{3}{4}+\frac{1}{4} \)

= 1

For 46. \( \sin ^{2} 225^{\circ}+\cos ^{2} 225^{\circ} \), we have:

= \( (-\frac{\sqrt{2}}{2})^{2}+(-\frac{\sqrt{2}}{2})^{2} \)

= \( \frac{2}{4}+\frac{2}{4} \)

= 1

For 47. \( 2 \tan ^{2} 120^{\circ}+3 \sin ^{2} 150^{\circ} \), we have:

= \( 2(\frac{\sqrt{3}}{3})^{2}+3(\frac{1}{2})^{2} \)

= \( 2(\frac{1}{3})+3(\frac{1}{4}) \)

= \( \frac{2}{3}+\frac{3}{4} \)

= \( \frac{8}{12}+\frac{9}{12} \)

= \( \frac{17}{12} \)

So the final answers are:

Example 4. 45: 1

46: 1

47: \( \frac{17}{12} \)

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Answer: VA: x = 2

Step-by-step explanation: In order to find the verticle asymtote you have to substitute values for x in order to get zero. Look in the denominator and see which value minus or plus the value next to it will equal zero. In this case, 2-2 is 0 so your verticle asymtote is 2. Keep in mind you want 0 in the denominator because that would mean the value on a graph is undefined, showing your asymtote.

Answer: 2

Step-by-step explanation:

The following function can be used to find b:

b= 7d

What is an equation?

An equation is a mathematical statement that proves two mathematical expressions are equal in algebra, and this is how it is most commonly used. In the equation 3x + 5 = 14, for instance, the two expressions 3x + 5 and 14 are separated.

7 boxes per day

=> For d days , the number  boxes will be 7d

total number of boxes

b= 7d

when b=  3,500 ,

3500= 7d

=> d= 3500/7 = 500 days

The following function can be used to find b:

b= 7d

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The second smallest number that has 1, 2, 3, 4, and 5 as factors is 360.

What is factorization?

A number or other mathematical object is factorization or factored when it is written as the product of numerous factors, often smaller or simpler things of the same sort.

To find the smallest number that has 1, 2, 3, 4, and 5 as factors, we can simply multiply these numbers together, since they are all factors of their product:

1 × 2 × 3 × 4 × 5 = 120

Now we need to find the second smallest number with these factors. One way to approach this is to list out the multiples of 120 until we find a number that also has the factors 1, 2, 3, 4, and 5, and is larger than 120.

Multiplying 120 by 2, 3, 4, 5, and 6 gives us the multiples:

240, 360, 480, 600, 720

Checking each of these multiples, we see that only 360 has all the factors 1, 2, 3, 4, and 5. Therefore, the second smallest number that has 1, 2, 3, 4, and 5 as factors is 360.

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The actual area of the lake 64000 square meters.

When two numbers are compared, the ratio between them shows how often the first number contains the second. As an illustration, the ratio of oranges to lemons in a dish of fruit is 8:6 if there are 8 oranges and 6 lemons present.

In this given question, we are first of all going to use the given ratio of the on map measurements to the actual measurements, that is 1:20000 to calculate the actual measurements as follows:

[tex]1.6cm^2=1\times 1.6cm^2[/tex]------->1

Using the ratio 1:20000 in 1.1, we get,

[tex]1.6\times1 cm^2= 1.6\times (20000)^2 cm^2 = 1.6\times 400000000 cm^2 = 640000000cm^2[/tex]-----> 2

As the actual measurement of the area of the lake.

So, 1m = 100 cm

=> [tex](1m)^2=(100cm)^2[/tex]

=> [tex]1m^2 = 10000cm^2[/tex]-------> 3

So, using equation 3 in the value obtained 2, we get,

=> [tex]640000000cm^2 = \frac{640000000}{10000} = 64000m^2[/tex]

Hence the actual area of the lake is 64000 square meters.

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The measures of the two interior angles are:

x = 62°

y = 118°

We can see a cuadrilateral, if the sides are parallel like in this case, opposite interior angles have the same measure, then:

x = 62°

And we know that adjacent angles should add up to 180°, then:

y+ 62° = 180°

y = 180° - 62° = 118°

These are the measures of the two interior angles.

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Answer:

Smaller number = 19

Larger number = 71

x + y = 90

x = 14 + 3y

Step-by-step explanation:

Let y represent the smaller number, and then represent the larger number as x, in terms of y:

Smaller number = y

Larger number = x = 3y + 14

Since the sum of the two numbers is 90, form an equation by adding them together in this notation:

y + 3y + 14 = 90

Simplify the equation:

4y + 14 = 90

4y = 76

y = 19

Therefore the smaller number is 19. Now substitute into the expression for the larger number to find its value:

x = 3(19) + 14 = 71

Now verify the two values sum to 90 like we expect:

19 + 71 = 90

Now we can use what we know to complete the equations, if x = 71:

x + y = 90 (we know they sum to 90)

x = 14 + 3y (as respresented above - multiplying by 3 and adding 14)

The solution to the system of equations is (0,1).

To solve the system of equations using elimination, we need to eliminate one of the variables. In this case, we will eliminate x by multiplying the first equation by -2 and the second equation by 6.

First equation: 6x+3y=3
Second equation: 2x+7y=7

Multiply the first equation by -2: -12x-6y=-6
Multiply the second equation by 6: 12x+42y=42

Now we can add the two equations together to eliminate x:
-12x-6y=-6
+12x+42y=42
_______________
0x+36y=36

Now we can solve for y:
36y=36
y=1

Now we can plug y back into one of the original equations to solve for x:
6x+3(1)=3
6x+3=3
6x=0
x=0

So the solution to the system of equations is (0,1).

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The quadratic equations are solved and the value of x are 1 and -7 respectively

A quadratic equation is a second-order polynomial equation in a single variable x , ax² + bx + c=0. with a ≠ 0. Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has at least one solution. The solution may be real or complex.

The roots of the quadratic equations are

x = [ -b ± √ ( b² - 4ac ) ] / ( 2a )

where ( b² - 4ac ) is the discriminant

when ( b² - 4ac ) is positive, we get two real solutions

when discriminant is zero we get just one real solution (both answers are the same)

when discriminant is negative we get a pair of complex solutions

Given data ,

Let the quadratic equation be represented as A

Now , the value of A is

a)

x² + 6x = 7

Adding 9 on both sides of the equation , we get

x² + 6x + 9 = 16

On simplifying the equation , we get

( x + 3 )² = ( 4 )²

Taking square roots on both sides , we get

x + 3 = ±4

Subtracting 3 on both sides , we get

x = 1 and x = -7

Therefore , the value of x are 1 and -7 respectively

Hence , the quadratic equations is solved

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2x+7y+2z+10=0 is the equation of the plane passing through A(-2, 0, -3) and B(1, -2, 1) that lies parallel to the line through C(-2, -13/5, 26/5) and D(16/5, -13/5, 0).

The equation of the plane passing through a point (a,b,c) can be written as A(x-a) + B(y-b) + C(z-c) = 0, where A, B and C are the coefficients of the normal vector to the plane.

So, the equation of the plane passing through a point (-2,0,-3) can be written as A(x+2) + B(y) + C(z+3) = 0,...i)

Now the plane also passes through the point (1,-2,1) so A(1+2) + B(-2) + C(1+3) = 0,

So, 3A-2B+4C=0.........ii)

Now, the direction cosines of CD is

l= 16/5 +2= 26/5

m= -13/5+13/5 = 0

n= 0-26/5 = -26/5

For a plane and line to be perpendicular Dot product of the direction cosines must be zero

or A*26/5 +  B*0 + C*-26/5=0

or, A=C.....iii)

Putting this in i) 7A-2B=0 or, A=2B/7....iv)

putting iii) and iv)

A(x+2) + B(y) + C(z+3) = 0

or, A(x+2) + B(y) + A(z+3) = 0

or, 2*B*(x+2)/7 + B(y) + 2*B*(z+3) /7= 0

or 2*(x+2)/7 + y + 2*(z+3) /7= 0

or 2x+7y+2z+10=0

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This is our final answer in terms of n.

The given sum is n Σ (3k – 3) = ____ , where k=1. To express this sum in closed form, we can use the formula for the sum of an arithmetic series. The formula is:

S = n/2 (a1 + an)

where S is the sum, n is the number of terms, a1 is the first term, and an is the last term.

In this case, the first term is 3(1) - 3 = 0 and the last term is 3(n) - 3 = 3n - 3. Plugging these values into the formula, we get:

S = n/2 (0 + 3n - 3)

Simplifying the equation gives us:

S = n/2 (3n - 3)

S = (3n^2 - 3n)/2

S = (3n^2)/2 - (3n)/2

S = (3/2)n^2 - (3/2)n

Therefore, the sum in closed form is:

n Σ (3k – 3) = (3/2)n^2 - (3/2)n

This is our final answer in terms of n.

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Solutions to Exercises 51-56:

The equation can be rewritten as: \( \cos x (2 \sin x-1)=0 \). The solutions are: \( \cos x=0 \) or \( \sin x=\frac{1}{2} \). The solutions in the interval \( [0,2 \pi) \) are: \( x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{\pi}{6}, \frac{5 \pi}{6} \).

The equation can be rewritten as: \( \tan x (\sqrt{2} \cos x-1)=0 \). The solutions are: \( \tan x=0 \) or \( \cos x=\frac{1}{\sqrt{2}} \). The solutions in the interval \( [0,2 \pi) \) are: \( x=0, \pi, \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \).

The equation can be rewritten as: \( \sin x (1-\cos x)=0 \). The solutions are: \( \sin x=0 \) or \( \cos x=1 \). The solutions in the interval \( [0,2 \pi) \) are: \( x=0, \pi, 2 \pi \).

Overall, the solutions to these exercises can be found by factoring the equations and finding the solutions to each factor in the given interval.

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The solutions in the interval \( [0,2 \pi) \) are x = 0, π, and 2π.

In Exercises 51-56, find all solutions to the equation in the interval \( [0,2 \pi) \). You do not need a calculator.

51. \( 2 \cos x \sin x-\cos x=0 \): The solutions in the interval \( [0,2 \pi) \) are x = 0, π, and 2π.

52. \( \sqrt{2} \tan x \cos x-\tan x=0 \): The solutions in the interval \( [0,2 \pi) \) are x = π/4, 3π/4, 5π/4, and 7π/4.

53. \( 2 \cos^2 x-\sin^2 x=1 \): The solutions in the interval \( [0,2 \pi) \) are x = 0, π, and 2π.

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The trigonometric ratios of sine and cosines indicates that we get;

Trigonometric ratios specify the relationship between two sides of a right triangle and the acute angle in the right triangle.

The specified equations are presented as follows;

1. sin(55) = cos(θ)

The sine of an angle is the ratio of the opposite side to the angle in a right to the hypotenuse side of the right triangle

The trigonometric ratio of the cosine of an angle is the ratio of the adjacent side to the angle to the hypotenuse side in a right triangle

The acute angles in a right triangle are complementary

Let A and B represent the acute angles of a right triangle, we get;

The adjacent side to the angle A is the opposite side to the angle B

The hypotenuse side remains the same, therefore;

cos(A) = sin(B), and sin(A) = cos(A)

Which indicates; sin(55) = cos(90 - 55) = cos(35)

sin(55) = cos(θ) = cos(35)

2. cos (28) = sin(90 - 28) = sin(62)

cos(28) = sin(62)

sin(θ) = sin(62) = cos(28)

3. cos(y) = sin(θ)

cos(y) = sin(90 - y)

cos(y) = sin(90 - y) = sin(θ)

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Fraction in simplest form is (3)/(4).

To add or subtract fractions with the same denominator, you simply add or subtract the numerators and keep the same denominator. Then, simplify the fraction if possible.

In this case, the fractions have the same denominator of 8, so we can simply add the numerators:

(1)/(8) + (5)/(8) = (1 + 5)/(8) = (6)/(8)

Now, we need to simplify the fraction by finding the greatest common factor (GCF) of the numerator and denominator. The GCF of 6 and 8 is 2, so we can divide both the numerator and denominator by 2 to get:

(6)/(8) = (6/2)/(8/2) = (3)/(4)

So the final answer is (3)/(4).

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Answer:

CD = 22

Step-by-step explanation:

You would double the side lengths given since this is a rectangle and set them equal to 80 cause that is the perimeter. So the equation would be 6z + 6 + 8z + 4 = 80. You would use the PEDMAS rule and subtract 6 from both sides which gives you:

6z + 8z +4 = 74

Subtract 4 from both sides

6z + 8z = 70

Add 6z + 8z

14z = 70

Divide 70 and 14z

z = 5.

AB and CD is equal in length so all you have to do is plug in 5 to z in the AB equation.

4(5) + 2

= 22

(1) The distance from the starting point to the final point is 589.5 km and  the bearing from the starting point to the final point is 16.7° north of east

(2) The distance from the ship to the port is 126.49 km and the bearing of the port from the ship is  18.4° south of west.

To solve this problem, we can use the law of cosines to find the distance from the starting point to the final point. We can also use trigonometry to find the bearing (angle) between the starting point and final point.

See the diagram attached:

We want to find the distance and bearing from the starting point (x) to the final point.

Using the law of cosines, we have:

c² = a² + b² - 2abcos(C)

c² = 500² + 150² - 2x500x150xcos(120°)

c² = 250,000 + 22,500 + 75,000

c² = 347,500

c = √347,500

c = 589.5 km

To find the bearing, we can use trigonometry. Let θ be the angle between the line from the starting point to the final point and due east.

tan(θ) = (150 km) / (500 km)

θ = tan⁻¹(0.3)

θ = 16.7°

2. To solve this problem, we can use the Pythagorean theorem to find the distance from the ship to the port. We can also use trigonometry to find the bearing (angle) between the ship and the port.

We want to find the distance and bearing of the port (P) from the ship (S).

Using the Pythagorean theorem, we have:

d² = 120² + 40²

d² = 14400 + 1600

d² = 16000

d = √(16000)

d = 126.49 km

To find the bearing, we can use trigonometry. Let θ be the angle between the line from the ship to the port and due west.

tan(θ) = (40 km) / (120 km)

θ = tan⁻¹ (1/3)

θ = 18.4°

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The maximum height reached by the baII is 9 meters.

In mathematics, a functiοn is a relatiοn between twο sets, typically called the dοmain and range, that assigns tο each element οf the dοmain a unique element οf the range. In οther wοrds, a functiοn is a rule οr a set οf rules that assοciates each input value with exactly οne οutput value.

Tο find the maximum height reached by the ball, we need to find the vertex of the parabolic functiοn [tex]h = -t^2 + 6t[/tex]. The vertex of a parabola in the form[tex]y = ax^2 + bx + c[/tex] is located at the point [tex](-b/2a, c - b^2/4a)[/tex].

In this case, the functiοn is[tex]h = -t^2 + 6t[/tex]t, which has a=-1, b=6, and c=0. Therefοre, the vertex οf the parabοla is located at:

t = -b/2a = -6/(-2) = 3

Tο find the maximum height, we substitute t = 3 into the functiοn:

[tex]h = -t^2 + 6t = -3^2 + 6(3) = 9 meters[/tex]

Therefοre, the maximum height reached by the baIl is 9 meters.

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Answer:

y = 129°

Step-by-step explanation:

the figure has opposite sides congruent and is therefore a parallelogram.

• consecutive angles are supplementary , that is

y + 51° = 180° ( subtract 51° from both sides )

y = 129°

The difference in interest earned between the twο investments is R5 118.08.

There are twο main types οf interest: simple interest and cοmpοund interest. Simple interest is calculated οnly οn the principal amοunt, while cοmpοund interest is calculated οn bοth the principal amοunt and any accumulated interest frοm previοus periοds.

Interest rates can be fixed οr variable. A fixed interest rate remains the same thrοughοut the lοan οr investment periοd, while a variable interest rate can change οver time depending οn market cοnditiοns.

Interest is an impοrtant cοncept in finance and ecοnοmics, and it plays a key rοle in bοrrοwing and lending, investments, and savings.

Investment A:

Simple interest = P × r × t

Where P = R5 500, r = 8% = 0.08 and t = 4 years

Simple interest = 5500 × 0.08 × 4

Simple interest = 1760

Investment B:

Cοmpοund interest = [tex]$A = P(1 + \frac{r}{n})^{nt}[/tex]

Where P = R5 500, r = 7.5% = 0.075, n = 1 (as it is cοmpοunded annually) and t = 3 years

Cοmpοund interest = [tex]$5500 \times (1 + \frac{0.075}{1} )^{(1 \times 3)} - 5500[/tex]

Cοmpοund interest = 1332.63

The difference between the interest earned οn the twο investments is:

Difference = Investment A - Investment B

Difference = 1760 - 1332.63

Difference = 427.37

Therefοre, the difference in interest earned between the twο investments is R5 118.08.

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Answer:

0.011

Step-by-step explanation:

Answer:9.05

Step-by-step explanation: uh why? its easy