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kherson [118]
4 years ago
15

Use the distributive property to remove the parentheses. -5(6y - u - 2)

Mathematics
1 answer:
Advocard [28]4 years ago
5 0

Answer:

-30y+5u+10

Hope this helped

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What is the range of the function f(x) = 4x + 9, given the domain D = {-4, -2, 0, 2}?
Hitman42 [59]

Answer:

Range D = {-7, 1, 9, 17}

Step-by-step explanation:

f(x) = 4x + 9

4(-4) + 9 = -7

4(-2) + 9 = 1

4(0) + 9 = 9

4(2) + 9 = 17

5 0
3 years ago
Please help! It’s so confusing
sammy [17]
I believe the answer is b:)
3 0
3 years ago
A blueprint has a scale of 2 inches = 2 feet. What is the scale factor of the blueprint?
VARVARA [1.3K]
1 in = 1 ft 

there are 12 inches in a foot...

you could say that 1 inch = 12 inches, or that the scale is 1:12
3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cbegin%7Bequation%7D%5Ctext%20%7B%20Question%3A%20If%20%7D%20%5Cln%20%28x%2By%29%3D4%20%5Cti
Art [367]

I think you meant to say

\ln(x+y) = 4xy

and not "4 times y" on the right side (which would lead to a complex value for y when x = 0). Note that when x = 0, the equation reduces to ln(y) = 0, so that y = 1.

Implicitly differentiating both sides with respect to x, taking y = y(x), and solving for dy/dx gives

\dfrac{1+\frac{dy}{dx}}{x+y} = 4y + 4x\dfrac{dy}{dx}

\implies \dfrac{dy}{dx} = \dfrac{4xy+4y^2-1}{1-4x^2-4xy}

Note that when x = 0 and y = 1, we have dy/dx = 3.

Differentiate both sides again with respect to x :

\dfrac{d^2y}{dx^2} = \dfrac{(1-4x^2-4xy)\left(4y+4x\frac{dy}{dx}+8y\frac{dy}{dx}\right)-(4xy+4y^2-1)\left(-8x-4y-4x\frac{dy}{dx}\right)}{(1-4x^2-4xy)^2}

No need to simplify; just plug in x = 0, y = 1, and dy/dx = 3 to get

\dfrac{d^2y}{dx^2} \bigg|_{x=0} = \boxed{40}

8 0
2 years ago
It is possible to construct a cube of twice the volume of a given cube using only a straightedge and compass.
Oxana [17]

No, it is not.

That specific question is named the Delian Problem, where supposedly, Apollo, the Greek god, had asked for his altar to be doubled in size to stop a plague going around. The builders only had a compass and straightedge, but they couldn't figure out how to double the volume. The reason they couldn't was because there was no way to construct multiples of \sqrt[3]{2a}, which was what was needed to double the volume.

This problem is one called the impossible problems from antiquity, or problems the ancient Greeks could not solve with only a compass and a straightedge. Another example of one of these problems is how to trisect an angle.

8 0
4 years ago
Read 2 more answers
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