Can someone solve this ??
I need help ASAP

The missing values are 2 and 35 respectively. The points plotted have been attached below. The solution has been obtained by using equivalent ratio.

When we compare two ratios, they are said to be equivalent. To determine whether two or more ratios are equivalent, they can be compared to one another.

We are given that Bashir walked 10 miles in 4 hours and the table is of equivalent ratios.

The ratio comes out to be 2.5.

So, he will walk 5 miles in 2 hours.

Similarly, in 14 hours he will walk 35 miles.

The points have been plotted on the coordinate axes and has been attached below.

Hence, the missing values are 2 and 35 respectively.

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The quotient and remainder using long division for [tex](2x^(3)-6x^(2)+7x-11)/(2x^(2)+5)[/tex] are x-11/2 and 55/2x-11, respectively.

The quotient and remainder using long division for the given expression can be found by following these steps:

Step 1: Set up the long division as follows:
[tex]```2x^(2)+5 | 2x^(3)-6x^(2)+7x-11         |----------------------         |```[/tex]

Step 2: Divide the first term of the dividend (2x^(3)) by the first term of the divisor (2x^(2)) to get the first term of the quotient (x):
[tex]```2x^(2)+5 | 2x^(3)-6x^(2)+7x-11         |----------------------         | x```[/tex]

Step 3: Multiply the first term of the quotient (x) by the divisor (2x^(2)+5) and subtract the result from the dividend:
[tex]```2x^(2)+5 | 2x^(3)-6x^(2)+7x-11         |-(2x^(3)+5x)         |----------------------         |        -11x^(2)+7x-11```[/tex]

Step 4: Repeat steps 2 and 3 until the degree of the remainder is less than the degree of the divisor:
[tex]```2x^(2)+5 | 2x^(3)-6x^(2)+7x-11         |-(2x^(3)+5x)         |----------------------         | x     -11x^(2)+7x-11         |      -(-11x^(2)-55/2x)         |----------------------         |         55/2x-11```[/tex]

Step 5: The final result is the quotient and remainder:
[tex]```Quotient = x-11/2Remainder = 55/2x-11```[/tex]

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Another polynomial with the same factored form of (z-3)(z+5) but different values for z is 2z^(2)+4z-30.

To find another polynomial with the same factored form, we can simply multiply the given factored form by a constant. This will change the values of z, but the factored form will remain the same.

For example, let's multiply the factored form by 2:

2(z-3)(z+5) = 2z^(2)+4z-30

In general form, this polynomial is 2z^(2)+4z-30.

We can see that the factored form is still (z-3)(z+5), but the values of z have changed. In the original polynomial, z^(2)+2z-15, the values of z were 3 and -5. In the new polynomial, 2z^(2)+4z-30, the values of z are 3/2 and -5/2.

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The image of point a after reflection over the y axis followed by dilation with a scale factor of 3 centered at the origin is,

⇒ (6, 9)

A pair of numbers which describe the exact position of a point on a cartesian plane by using the horizontal and vertical lines is called the coordinates.

Given that;

The coordinates of point a are,

⇒ a =  (-2,3)

Now, We can multiply by 3 in the point;

⇒ (- 2, 3) = (- 2 × 3, 3×3)

              = (- 6, 9)

Now, After reflection over the y axis,

Image of the point are,

⇒ (6, 9)

Thus, The image of point a after reflection over the y axis followed by dilation with a scale factor of 3 centered at the origin is,

⇒ (6, 9)

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The intersection points between the two equations are (1, 3) and (-4, 2)

Here we have the system of equations:

(x+4)^2+(y−3)^2=25

y=x+2

We can replace the second equation into the first one to get:

(x+4)^2+(x + 2−3)^2 = 25

(x + 4)^2 + (x - 1)^2 = 25

Now let's solve that:

x^2 + 8x + 16 + x^2 - 2x + 1 = 25

2x^2 + 6x + 17 - 25 = 0

2x^2 + 6x - 8 = 0

x^2 + 3x - 4 = 0

Using the quadratic formula:

[tex]x = \frac{-3 \pm \sqrt{3^2 - 4*1*-4} }{2} \\\\x = \frac{-3 \pm5 }{2}[/tex]

So the two solutions are:

x = (-3 + 5)/2 = 1

x = (-3 - 5)/2 = -4

Evaluating the line in these values:

y = 1 + 2 = 3 ----> we have the point (1, 3)

y = -4 + 2 = -2 ----> we have the point (-4, -2)

These are the intersection points.

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There are infinitely many expressions equivalent to 11x + 10, including 22x + 20, 11(x+1)-1, -11(-x)-10, and 11(x+2)-12.

What is expression ?

In mathematics, an expression is a combination of numbers, symbols, and/or variables that are put together in a meaningful way, usually to represent a quantity or a mathematical relationship.

There are infinitely many expressions that are equivalent to 11x + 10, because you can add or subtract any expression to both sides of the equation to get a new equivalent expression. Here are some examples:

22x + 20: This is equivalent to 11x + 10 because if you distribute 11 to x and 10, you get 11x + 10.

11(x + 1) - 1: This is also equivalent to 11x + 10 because if you distribute 11 to x and 1, you get 11x + 11 - 1, which simplifies to 11x + 10.

-11(-x) - 10: This is equivalent to 11x + 10 because if you distribute -11 to -x, you get 11x + 10.

11(x + 2) - 12: This is also equivalent to 11x + 10 because if you distribute 11 to x and 2, you get 11x + 22 - 12, which simplifies to 11x + 10.

In general, any expression of the form 11x + k, where k is a constant, is equivalent to 11x + 10.

Therefore, there are infinitely many expressions equivalent to 11x + 10, including 22x + 20, 11(x+1)-1, -11(-x)-10, and 11(x+2)-12.

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Answer:

14.1 feet

Step-by-step explanation:

Amount of liquid gallons in 1 cubic foot= 7.48

Radius of the cylindrical water tank= 2.2 feet

Amount of liquid gallons it can hold= 1,600

Volume of the cylindrical water tank to carry 1,600 liquid gallons= [tex]\frac{1600}{7.48}[/tex]

                                                                                                           = 213.90 ft³

Volume of the cylinder= [tex]\pi r^{2} h[/tex]

                            213.9= 3.14 x 2.2 x 2.2 x h

                            213.9= 3.14 x 4.84 x h

                            213.9= 15.2 x h

                                  h= [tex]\frac{213.9}{15.2}[/tex]

                                  h= 14.07

                                  h= 14.1 feet

∴ height of the water tank is 14.1 feet.

The integer values of k such that [18]25 is equal to [7]k are 4, 11, 18, 25

To find all integer values of k such that [18]25 is equal to [7]k, we need to use the following steps:

Start with the given equation: [18]25 = [7]k

Use the definition of modular arithmetic to write the equation in a different form: 25 ≡ k (mod 7)

Simplify the equation: 4 ≡ k (mod 7)

Find all integer values of k that satisfy the equation. The smallest positive integer value of k that satisfies the equation is 4. We can add multiples of 7 to find other integer values of k: 4 + 7 = 11, 4 + 14 = 18, 4 + 21 = 25, etc.

The integer values of k that satisfy the equation are 4, 11, 18, 25, etc.

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Answer:

a,b,c,d

Step-by-step explanation:

all are rational except efg

The equation is true for n = k+1, so the statement is true for all n>=1 by mathematical induction.

Proof by mathematical induction:

Base case: n = 1
1^(3) = ((1(1+1))/(2))^(2)
1 = ((2)/(2))^(2)
1 = 1^(2)
1 = 1
The base case is true.

Inductive step:
Assume that the statement is true for n = k, that is:
1^(3)+2^(3)+3^(3)+...+k^(3) = ((k(k+1))/(2))^(2)

Now we need to prove that the statement is also true for n = k+1:
1^(3)+2^(3)+3^(3)+...+k^(3)+(k+1)^(3) = (((k+1)((k+1)+1))/(2))^(2)

Substituting the assumption into the left-hand side of the equation:
((k(k+1))/(2))^(2) + (k+1)^(3) = (((k+1)((k+1)+1))/(2))^(2)

Expanding the right-hand side of the equation:
((k(k+1))/(2))^(2) + (k+1)^(3) = (((k+1)(k+2))/(2))^(2)

Simplifying the equation:
(k^(2)(k+1)^(2))/(2^(2)) + (k+1)^(3) = ((k+1)^(2)(k+2)^(2))/(2^(2))

Multiplying both sides of the equation by 2^(2):
(k^(2)(k+1)^(2)) + 2^(2)(k+1)^(3) = (k+1)^(2)(k+2)^(2)

Expanding the equation:
k^(2)(k+1)^(2) + 2^(2)(k+1)^(3) = (k+1)^(2)(k^(2)+4k+4)

Simplifying the equation:
k^(2)(k+1)^(2) + 2^(2)(k+1)^(3) = k^(2)(k+1)^(2) + 4k(k+1)^(2) + 4(k+1)^(2)

Subtracting k^(2)(k+1)^(2) from both sides of the equation:
2^(2)(k+1)^(3) = 4k(k+1)^(2) + 4(k+1)^(2)

Factoring out (k+1)^(2) from the right-hand side of the equation:
2^(2)(k+1)^(3) = (k+1)^(2)(4k+4)

Simplifying the equation:
2^(2)(k+1)^(3) = 4(k+1)^(2)(k+1)

Dividing both sides of the equation by (k+1)^(2):
2^(2)(k+1) = 4(k+1)

Simplifying the equation:
2^(2)(k+1) = 2^(2)(k+1)

The equation is true for n = k+1, so the statement is true for all n>=1 by mathematical induction.

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The answer to the question is that \( A_{1} \) is smaller than \( A_{2} \)

To begin, it is important to note that the terms "emalier" and "5PRECALC7" are not relevant to the question and can be ignored. Additionally, there are several typos and extraneous information that can also be ignored. The main focus of the question is to determine the relationship between \( A_{1} \) and \( A_{2} \).

From the information provided, it is clear that \( A_{1} \) is smaller than \( A_{2} \). This is because the value of \( A_{1} \) is given as 2, while the values of \( b \) and \( c \) are 27 and 33, respectively. Since \( A_{2} \) is the sum of \( b \) and \( c \), it is clear that \( A_{2} \) is larger than \( A_{1} \).

Therefore, the answer to the question is that \( A_{1} \) is smaller than \( A_{2} \).

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The correct factorization for the Difference of Squares or a²-b² is (a+b)(a-b). This is because when you multiply (a+b)(a-b), you get a²- ab + ab - b², which simplifies to a²-b².

Here is a step-by-step explanation of how to factor the Difference between Squares:

Step 1: Identify the two terms that are being squared. In this case, a and b are the terms being squared.
Step 2: Write the two terms with a plus sign in between them in one set of parentheses and with a minus sign in between them in another set of parentheses. This will give you (a+b)(a-b).
Step 3: Multiply the two sets of parentheses together to check your answer. You should get a²-b².
So, the correct factorization for the Difference of Squares or a²-b² is (a+b)(a-b).

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Answer:

Below

Step-by-step explanation:

Diameter = 20 inches     radius = 10 inches

Area = pi r^2 = 3.14 * 10^2 = 314 in^2

An estimated 46 students in both surveys combined chose in-line skating as their favorite.

To estimate how many students in both surveys combined chose in-line skating as their favorite, we can use the given percentage and the total number of students surveyed.
First, we need to calculate the number of students who chose in-line skating in the first survey. We can do this by multiplying the percentage by the total number of students surveyed:
23% × 100 = 23
So, 23 students in the first survey chose in-line skating as their favorite.
Next, we need to add this number to the number of students who chose in-line skating in the second survey. Since we don't have information about the second survey, we can assume that the same percentage of students chose in-line skating:
23% × 100 = 23
Finally, we can add the two numbers together to get the total number of students who chose in-line skating in both surveys:
23 + 23 = 46

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Answer:

Kayla and Rachel raised $8.39, Francesca raised $18.89.

Step-by-step explanation:

Knowing that $35.67 is the total amount raised, that Rachel and Kayla raised the same amount (x) each, and that Francesca raised $10.50 more than them (x + 10.50), then the equation would look like:

35.67 = x + x + (x + 10.50)

or

35.67 = 3x + 10.50

From here, you can solve algebraically.

Isolate the variable by subtracting 10.50 from both sides

25.17 = 3x

Then, find X by dividing both sides by 3:

x = 8.39

Knowing that Kayla and Rachel raised $8.39 each, to find Francesca's contribution you can either:

A) Subtract Kayla and Rachel's contributions from the total amount:

     35.67 - 8.39 - 8.39 = 18.89

or

B) Add $10.50 to the amount that Kayla and Rachel raised:

     8.39 + 10.50 = 18.89

Step-by-step explanation:

[tex]18 + \frac{7}{10} - 16 + \frac{26}{6}\\ 30(18) + 30 \times \frac{7}{10} - 30(16) + 30 \times \frac{26}{6} \\ 540 + 21 - 480 + 130 \\ = 211[/tex]

The Lcm of 10 and 6

is 30

sorry but I got my answer to be 211

Answer:

this is right.....

Step-by-step explanation:

15x + 6

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The simplified root of the given root i.e. √(400x¹⁰⁰) is 20x⁵⁰. The solution has been obtained by using the law of indices.

What is the law of indices?

The guidelines for simplifying expressions containing powers of the same base number are known as index laws.

We are given an expression as √(400x¹⁰⁰).

Using law of indices,

⇒ √(400x¹⁰⁰) = √400 * √(x¹⁰⁰)             ( as √ab = √a * √b)

⇒ √(400x¹⁰⁰) = 20 √(x¹⁰⁰)

⇒ √(400x¹⁰⁰) = 20 √(x⁵⁰ * x⁵⁰)             ( as a⁴ = a² + a²)

⇒ √(400x¹⁰⁰) = 20x⁵⁰

Hence, the simplified root of the given root i.e. √(400x¹⁰⁰) is 20x⁵⁰.

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The answer of third integer is 15.

To solve this problem, we can use algebra. Let's call the first of the three consecutive odd integers "x." The next two consecutive odd integers would then be "x + 2" and "x + 4."

The problem tells us that three times the first integer is 3 more than twice the third, so we can write an equation:

3x = 2(x + 4) + 3

Simplifying the equation, we get:

3x = 2x + 8 + 3

x = 11

So the first integer is 11, and the third integer is 11 + 4 = 15.

Therefore, the correct answer is O 15.

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Answer:

The volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius.

Substituting r = 15, we get:

V = (4/3)π(15)^3

V = (4/3)π(3375)

V = 4π(1125)

V = 4500π

Therefore, the volume of the sphere with a radius of 15 units is 4500π cubic units. This can also be approximated as 14,137.17 cubic units by using a value of 3.14 for π and rounding to the nearest hundredth.

Step-by-step explanation:

Answer:

I found this

Step-by-step explanation:

To simplify 3x² * 4x⁵, we can multiply the coefficients (numbers in front of the variables) and add the exponents of x:

3x² * 4x⁵ = (3 * 4) x^(2+5) = 12x^7

Therefore, the simplified expression is 12x^7.

The exponential function that passes through the two points given is (a) f(x) = 4(2ˣ) and (b) g(x) = (9^(2/5)) ((1/9)^(1/5))ˣ.

1. Use the general form of an exponential function: f(x) = abˣ
2. Plug in the x and y values from the first point into the equation and solve for b.
3. Plug in the x and y values from the second point and the value of b into the equation and solve for a.
4. Substitute the values of a and b back into the general equation to get the formula for the function.

For point a:

1. f(x) = abˣ
2. 2 = ab⁻¹
3. 64 = ab⁴
4. Divide the equations to eliminate a: (64/2) = (ab⁴)/(ab⁻¹) = b⁵
5. Solve for b: b = 2
6. Substitute b back into one of the equations and solve for a: 2 = a(2)⁻¹ = a/2, a = 4
7. Substitute a and b back into the general equation: f(x) = 4(2)ˣ

For point b:

1. g(x) = abˣ
2. 9 = ab⁻³
3. 1 = ab²
4. Divide the equations to eliminate a: (1/9) = (ab²)/(ab⁻³) = b⁵
5. Solve for b: b = (1/9)^(1/5)
6. Substitute b back into one of the equations and solve for a: 9 = a((1/9)^(1/5))⁻³ = a(9^(3/5)), a = 9^(2/5)
7. Substitute a and b back into the general equation: g(x) = (9^(2/5))((1/9)^(1/5))ˣ

So the formulas for the exponential functions are:
f(x) = 4(2)ˣ
g(x) = (9^(2/5)) ((1/9)^(1/5))ˣ

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Answer: X = 14

Step-by-step explanation: (:

Answer:

x=2

Step-by-step explanation:

2(5x+3)=26

expand the bracket first

10x+6=26

take away 6 from both sides

10x=20

divided by 10 on each side

x=2

Answer:

You would use the PEDMAS rule for this equation.

First, distribute 2 inside the parentheses so the equation is 10x + 6 = 26

Then, subtract 6 from both sides 10x = 20

The answer is x = 2.

Step-by-step explanation:

30 peοple are playing chess and 30 peοple are playing cribbage.

A system οf linear equatiοns is a set οf twο οr mοre linear equatiοns with multiple variables that are sοlved simultaneοusly. The gοal οf sοlving a system οf linear equatiοns is tο find the values οf the variables that satisfy all the equatiοns in the system.

A linear equatiοn is an equatiοn in which the variables appear οnly tο the first pοwer, and the cοefficients οf the variables are cοnstants. Fοr example, the equatiοn y = 3x + 2 is a linear equatiοn, where y and x are the variables, and 3 and 2 are the cοefficients.

A system οf linear equatiοns can have οne unique sοlutiοn, nο sοlutiοn, οr infinitely many sοlutiοns. The number οf sοlutiοns depends οn the number οf equatiοns and variables in the system, and the relatiοnships between them. There are variοus methοds tο sοlve a system οf linear equatiοns, such as substitutiοn, eliminatiοn, and matrix methοds.

Let's assume that there are x peοple playing chess and y peοple playing cribbage. Since everyοne is playing either chess οr cribbage, we knοw that:

x + y = 60

We alsο knοw that chess is a twο-player game, while cribbage is a fοur-player game. Therefοre, the tοtal number οf players in the games must be a multiple οf 2 and 4. This means that:

2x + 4y = 60

We can simplify this equatiοn by dividing bοth sides by 2

x + 2y = 30

Nοw we have twο equatiοns:

x + y = 60

x + 2y = 30

We can sοlve fοr x by subtracting the secοnd equatiοn frοm the first equatiοn:

x + y - (x + 2y) = 60 - 30

-y = -30

y = 30

Substituting this value οf y intο the first equatiοn, we get:

x + 30 = 60

x = 30

Therefοre, 30 peοple are playing chess and 30 peοple are playing cribbage.

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Complete question:

60 people attend a game night. Everyone chooses to play chess, a two-player game, or cribbage, a four-player game. All 60 people are playing either chess or cribbage. There are 3 more games of cribbage being played than games of chess being played. How many of each game are being played?

8 chess, 15 cribbage

30 chess, 30 cribbage

8 chess, 11 cribbage

30 chess, 15 cribbage

It wiII cοst Jοnah $63.36 tο decοrate the 3-tiered circuIar wedding cake with fresh red rοses arοund the base οf each IeveI.

Tο caIcuIate the cοst οf decοrating the cake with rοses, we need tο first determine the circumference οf each tier, which wiII teII us hοw many rοses are needed fοr each tier.

The circumference οf a circIe can be caIcuIated using the fοrmuIa C = πd, where C is the circumference, π is the mathematicaI cοnstant pi (apprοximateIy 3.14), and d is the diameter.

Fοr the bοttοm tier with a diameter οf 14 inches, the circumference is C = πd = 3.14 x 14 = 43.96 inches.

Fοr the middIe tier with a diameter οf 10 inches, the circumference is C = πd = 3.14 x 10 = 31.4 inches.

Fοr the tοp tier with a diameter οf 6 inches, the circumference is C = πd = 3.14 x 6 = 18.84 inches.

Nοw, we need tο determine hοw many rοses are needed fοr each tier. Tο dο this, we need tο caIcuIate hοw many rοses wiII fit arοund the circumference οf each tier. We can dο this by dividing the circumference οf each tier by the width οf the rοses, which is 1.5 inches.

Fοr the bοttοm tier: 43.96 inches / 1.5 inches per rοse = 29.3, rοunded up tο 30 rοses.

Fοr the middIe tier: 31.4 inches / 1.5 inches per rοse = 20.9, rοunded up tο 21 rοses.

Fοr the tοp tier: 18.84 inches / 1.5 inches per rοse = 12.6, rοunded up tο 13 rοses.

Nοw that we knοw hοw many rοses are needed fοr each tier, we can caIcuIate the tοtaI cοst οf the rοses. Each rοse cοsts $0.99, sο the cοst οf the rοses fοr each tier is:

Bοttοm tier: 30 rοses x $0.99 per rοse = $29.70

MiddIe tier: 21 rοses x $0.99 per rοse = $20.79

Tοp tier: 13 rοses x $0.99 per rοse = $12.87

The tοtaI cοst οf decοrating the cake with rοses is the sum οf the cοsts fοr each tier:

$29.70 + $20.79 + $12.87 = $63.36

Therefοre, it wiII cοst Jοnah $63.36 tο decοrate the 3-tiered circuIar wedding cake with fresh red rοses arοund the base οf each IeveI.

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The system of equations 4x + 2y = 6 and y = -2x + 6  have infinite many solutions

An equation is an expression that shows the relationship between two or more numbers and variables. Equations can either be linear, quadratic, cubic and so on depending on the degree.

Given the system of equation:

4x + 2y = 6

Divide through by 2:

2x + y = 3

y = -2x + 6     (1)

The second equation is:

y = -2x + 6      (2)

Since both equations are the same hence the system of equations have infinite many solutions

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Since TB = TS and OT is perpendicular to RS, it follows that BO is also perpendicular to RS, and hence RS is the diameter of the circle.

The circle is defined as the locus of the point traces around a fixed point called the centre and is equidistant from the out trace.

Since T is the midpoint of both chords PQ and RS, it follows that the perpendicular bisectors of these chords pass through the centre of the circle. Let O be the centre of the circle.

Consider the perpendicular bisector of PQ passing through T. Let it intersect the circle at point A. Since TA = TP and OT is perpendicular to PQ, it follows that AO is also perpendicular to PQ, and hence PQ is the diameter of the circle.

Similarly, consider the perpendicular bisector of RS passing through T. Let it intersect the circle at point B. Since TB = TS and OT is perpendicular to RS, it follows that BO is also perpendicular to RS, and hence RS is the diameter of the circle.

Therefore, PQ and RS are both diameters of the circle.

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