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3 years ago

11

128 is the product of Julie's savings and 8 Use the variable j to represent Julie's savings.

1 answer:

Readme [11.4K]3 years ago

4 0

128=8j; divide both sides by 8

j=16

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A = 1/2bh, where A is the surface area and b is the base, and H is the height.

elena55 [62]

When height turns to 3 times height or 3h

we can replace h with 3h and see what happens

A=1/2b(3h)
A=3/2bh
A=3(1/2bh)
well, the area is trippled

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3 years ago

The following is the converse of a conditional statement. If it is not April 17th, then it is not Matt's birthday. What is the i

user100 [1]

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3 years ago

Which best explains why |-4| = 4?

Dmitry [639]

Answer:

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2 years ago

What is the solution to the system of linear equations?

AlexFokin [52]

X=8-y

2(8-y)=10
16-2y=10
-2y=10-16
-2y=-6
y=3

x+3=8
x=8-3
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(5,3) is the answer :)

8 0

2 years ago

The product of (4z^2 + 7z - 8) and (-z + 3) is -4z^3 + xz^2 + yz - 24. What is the value of x? What is the value of y?

ycow [4]

Answer:

$x=5+\frac{29}{z}-\frac{y}{z}$

$y=5z+29-xz$

Step-by-step explanation:

Given: $(4z^2+7z-8)(-z+3)=-4z^3+xz^2+yz-24$ ; find x and y.

Let's start with finding x first. Let's distribute the left side of the equation.

$-4z^3+5z^2+29z-24=-4z^3+xz^2+yz-24$

Let's move everything we can to the left side of the equation. Some things will cancel out.

$5z^2+29z=xz^2+yz$

Now, let's move the term that includes $x$ to the left side of the equation. Everything else should be moved to the right.

$-xz^2=-5z^2-29z+yz$

Let's get rid of the negative sign in front of $x$ by dividing both sides by -1.

$xz^2=5z^2+29z-yz$

To get $x$ completely isolated, divide both sides by $z^2$ .

$\frac{xz^2}{z^2}=\frac{5z^2}{z^2}+\frac{29z}{z^2}-\frac{yz}{z^2}$

Simplify.

$x=5+\frac{29}{z}-\frac{y}{z}$

Let's solve for y. We will start with the distributed and simplified equation to make it easier.

$5z^2+29z=xz^2+yz$

Now, let's move the term that includes $y$ to the left side of the equation. Everything else should be moved to the right.

$-yz=-5z^2-29z+xz^2$

Again, let's get rid of the negative sign in front of the $y$ by dividing both sides by -1.

$yz=5z^2+29z-xz^2$

To isolate y, divide both sides by z.

$\frac{yz}{z}=\frac{5z^2}{z}+\frac{29z}{z}-\frac{xz^2}{z}$

Simplify.

$y=5z+29-xz$

8 0

2 years ago

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