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Sonbull [250]
2 years ago
8

Your boss sends you a text of a painting of the Yankee stadium.

Mathematics
1 answer:
kaheart [24]2 years ago
4 0

Answer:

170.6 inches

The scale factor is an enlargement

Step-by-step explanation:

16 by 24 is the Painting by a scale factor of 2.25

A scale factor is an enlargment or reduction in a shape or quantity.

So

(16*24) is the size of the painting after Scale factor

16*24=384

Know get 384/2.25

170.666666667

170.6 inches

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Novosadov [1.4K]
7 / 1/3
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Elden [556K]

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The quotient is 10x + 16

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3 years ago
Consider the function represented by the equation 6q = 3s - 9. Write the equation in function notation, where q is the independe
serg [7]

Answer:

s = 2q + 3

Step-by-step explanation:

A linear function has the form:

● y = mx + b

● y is the output of the function

● x is the variabke that we input

● b is the y-intetcept.

Focus on y and x.

Notice that y depends of the value of x. The value of y changes by changing x. So the value of x controls the output y.

y is dependent but x is not.

■■■■■■■■■■■■■■■■■■■■■■■■■■

● 6q = 3s - 9

We want q to be the independent variable wich means that q will be the input. Therefore s should be the output.

The strategy we are going to follow is separating s in one side alone.

● 6q = 3s - 9

Add 9 to both sides

● 6q + 9 = 3s -9 + 9

● 6q + 9 = 3s

Divide both sides by 3

● (6q + 9)/3 = (3s)/3

● (6q)/3 + 9/3 = s

● s = 2q + 3

So the answer is s = 2q + 3

6 0
3 years ago
Given f(x) = (lnx)^3 find the line tangent to f at x = 3
kirill [66]
Explanation

We must the tangent line at x = 3 of the function:

f(x)=(\ln x)^3.

The tangent line is given by:

y=m*(x-h)+k.

Where:

• m is the slope of the tangent line of f(x) at x = h,

,

• k = f(h) is the value of the function at x = h.

In this case, we have h = 3.

1) First, we compute the derivative of f(x):

f^{\prime}(x)=\frac{d}{dx}((\ln x)^3)=3*(\ln x)^2*\frac{d}{dx}(\ln x)=3*(\ln x)^2*\frac{1}{x}=\frac{3(\ln x)^2}{x}.

2) By evaluating the result of f'(x) at x = h = 3, we get:

m=f^{\prime}(3)=\frac{3}{3}*(\ln3)^2=(\ln3)^2.

3) The value of k is:

k=f(3)=(\ln3)^3

4) Replacing the values of m, h and k in the general equation of the tangent line, we get:

y=(\ln3)^2*(x-3)+(\ln3)^3.

Plotting the function f(x) and the tangent line we verify that our result is correct:

Answer

The equation of the tangent line to f(x) and x = 3 is:

y=(\ln3)^2*(x-3)+(\ln3)^3

6 0
1 year ago
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