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3 years ago

A side of a square is 8 3/2 inches. Using the area formula A = s2, determine the area of the square.

1 answer:

4 0

We are given that:
side length = 8 3/2 inches = 9.5 inches
the rule for the area is: A = s^2

Substitute with the side length in the rule given to get the area as follows:
Area = s^2 = (9.5)^2 = 90.25 inch^2

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If f(x) = 3x - 7, then f-1(x)=

fgiga [73]

Answer:

Step-by-step explanation:

so f-1(x) means we are finding the inverse

f(x) = 3x - 7.....change f(x) to y

y = 3x - 7.....now switch ur x and y and solve for y

x = 3y - 7....add 7 to both sides

x + 7 = 3y...divide both sides by 3

(x/3) + 7/3 = y....rearrange

y = (x/3) + 7/3....can also be written as : y = 1/3x + 7

now just change ur y to f-1(x)

f-1(x) = x/3 + 7/3.....or f-1(x) = 1/3x + 7/3

4 0

3 years ago

Read 2 more answers

The circle passes through the point (-4,-1). What is its radius?

liubo4ka [24]

Answer:

c is the correct answer for this question

Step-by-step explanation:

c is the correct answer for this question

1.5

2.5

Hope this helps.

4 0

3 years ago

. A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius of the base and height) of the drum

Drupady [299]

Answer:

$r=1.05\ \text{ft}$

$h=2.12\ \text{ft}$

$20.91\ \text{ft}^3$

Step-by-step explanation:

r = Radius

h = Height

Volume of cylinder = $7.35\ \text{ft}^3$

$V=\pi r^2h\\\Rightarrow h=\dfrac{V}{\pi r^2}\\\Rightarrow h=\dfrac{7.35}{\pi r^2}$

Surface area is given by

$A=2\pi r^2+2\pi rh\\\Rightarrow A=2\pi r^2+2\pi r\dfrac{7.35}{\pi r^2}\\\Rightarrow A=2\pi r^2+\dfrac{14.7}{r}$

Differentiating with respect to radius we get

$\dfrac{dA}{dr}=4\pi r-\dfrac{14.7}{r^2}$

Equating with zero we get

$0=4\pi r-\dfrac{14.7}{r^2}\\\Rightarrow 4\pi r=\dfrac{14.7}{r^2}\\\Rightarrow r^3=\dfrac{14.7}{4\pi}\\\Rightarrow r=(\dfrac{14.7}{4\pi})^{\dfrac{1}{3}}\\\Rightarrow r=1.05$

$\dfrac{d^2A}{dr^2}=4\pi-2\times \dfrac{14.7}{r^3}\\\Rightarrow \dfrac{d^2A}{dr^2}=4\pi+2\times \dfrac{14.7}{1.05^3}\\\Rightarrow \dfrac{d^2A}{dr^2}=37.96>0$