Kerstin has a square tile.the perimeter of the tile is 32 inches.what is the length of each side of the tile

Please help me i dont know how to do this!

jekas [21]

Slope = 2/3
Y-intercept =(0,-1)
Hope this helps

6 0

3 years ago

Alaska has a land area of about 1700000 square miles Florida has a land area one tenth of the size of Alaska what is the land ar

AnnyKZ [126]

Answer:

The land area of Florida is $170,000\ mi^2$

Step-by-step explanation:

Let

x -----> the land area of Alaska in square miles

y -----> the land area of Florida in square miles

we know that

The linear equation that represent this problem is equal to

$y=\frac{1}{10}x$ ----> equation A

$x=1,700,000$ ----> equation B

substitute equation B in equation A and solve for y

$y=\frac{1}{10}(1,700,000)=170,000\ mi^2$

Remember that

Florida's land area is one-tenth the size of Alaska is the same as to say that Alaska's land area is ten times the size of Florida

8 0

3 years ago

Please help

Zanzabum

Answer:

The value of n = 12

Step-by-step explanation:

Given the points

(-4, n)

(2, 4n)

The slope formula

m = y₂-y₁ / x₂ - x₁

$\left(x_1,\:y_1\right)=\left(-4,\:n\right),\:\left(x_2,\:y_2\right)=\left(2,\:4n\right)$

$m=\frac{4n-n}{2-\left(-4\right)}$

$m=\frac{n}{2}$

Given that the slope = m = 6

Thus, substituting the value

6 = n/2

n = 6 × 2

= 12

Thus, the value of n = 12

5 0

3 years ago

(2x+1)(2x-1) find the product

VladimirAG [237]

Difference of 2 perfect squares
a^2-b^2=(a-b)(a+b)
recognize that

(2x+1)(2x-1)
(a+b)(a-b)
therefor
product is
(2x)^2-(1)^2=
4x^2-1

8 0

3 years ago

Need help with AP CAL

anzhelika [568]

Answer: Choice C

$\displaystyle \frac{1}{2}\left(1 - \frac{1}{e^2}\right)$

============================================================

Explanation:

The graph is shown below. The base of the 3D solid is the blue region. It spans from x = 0 to x = 1. It's also above the x axis, and below the curve $y = e^{-x}$

Think of the blue region as the floor of this weirdly shaped 3D room.

We're told that the cross sections are perpendicular to the x axis and each cross section is a square. The side length of each square is $e^{-x}$ where 0 < x < 1

Let's compute the area of each general cross section.

$\text{area} = (\text{side})^2\\\\\text{area} = (e^{-x})^2\\\\\text{area} = e^{-2x}\\\\$

We'll be integrating infinitely many of these infinitely thin square slabs to find the volume of the 3D shape. Think of it like stacking concrete blocks together, except the blocks are side by side (instead of on top of each other). Or you can think of it like a row of square books of varying sizes. The books are very very thin.

This is what we want to compute

$\displaystyle \int_{0}^{1}e^{-2x}dx\\\\$

Apply a u-substitution

u = -2x

du/dx = -2

du = -2dx

dx = du/(-2)

dx = -0.5du

Also, don't forget to change the limits of integration

If x = 0, then u = -2x = -2(0) = 0
If x = 1, then u = -2x = -2(1) = -2

This means,

$\displaystyle \int_{0}^{1}e^{-2x}dx = \int_{0}^{-2}e^{u}(-0.5du) = 0.5\int_{-2}^{0}e^{u}du\\\\\\$

I used the rule that $\displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$ which says swapping the limits of integration will have us swap the sign out front.

--------

Furthermore,

$\displaystyle 0.5\int_{-2}^{0}e^{u}du = \frac{1}{2}\left[e^u+C\right]_{-2}^{0}\\\\\\= \frac{1}{2}\left[(e^0+C)-(e^{-2}+C)\right]\\\\\\= \frac{1}{2}\left[1 - \frac{1}{e^2}\right]$

In short,

$\displaystyle \int_{0}^{1}e^{-2x}dx = \frac{1}{2}\left[1 - \frac{1}{e^2}\right]$

This points us to choice C as the final answer.

5 0

2 years ago