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3 years ago

What is the goal of solving equations?

2 answers:

6 0

The goal in solving an equation is to get the variable by itself on one side of the equation and a number on the other side of the equation. To isolate the variable, we must reverse the operations acting on the variable.

irina [24]3 years ago

6 0

So In the future you know how to do math cause you need math for everything basically in life

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Find the length of XV for Rectangle WXYZ.<br><br> PLEASE HELP!!!!

Kaylis [27]

Answer: 20.5

Step-by-step explanation:

5 0

2 years ago

PLEASE HELP

ollegr [7]

Answer:

The answer is c i think not 100% sure

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Area of rectangular base

Amiraneli [1.4K]

Answer:

60 in^2

Step-by-step explanation:

Formula A=LW

Substitution:

$A=6*10\\A=60$

Easy Pea-sy!

I mean, there it only one answer so...

P.S. Hope you do good on this Canvas Test!(Yes, I recognize the interface.)

6 0

3 years ago

Write the number with the same value as 32 tens

sukhopar [10]

30 tens + 2 ones = 32

Hope I helped :P

7 0

3 years ago

Read 2 more answers

A cereal company is exploring new cylinder-shaped containers. Two possible containers are shown in the diagram.x is 6 and r=4.5

KatRina [158]

Answer:

Container $\rm X$ will have less label area than container $\rm Y$ by about $9\; \rm in$ .

Step-by-step explanation:

A rectangular sheet of paper can be rolled into a cylinder. Conversely, the lateral surface of a cylinder can be unrolled into a rectangle- without changing the area of that surface.

Indeed, the width of that rectangle will be the same as the height of the cylinder. On the other hand, the length of that rectangle should be exactly equal to the circumference of the circles on the top and the bottom of the cylinder. In other words, if a cylinder has a height of $h$ and a radius of $r$ at the top and the bottom, then its lateral surface can be unrolled into a rectangle of width $h$ and length $2\,\pi\, r$ .

Apply this reasoning to both cylinder $\mathrm{X}$ and $\rm Y$ :

For cylinder $\mathrm{X}$ , $h = 6\; \rm in$ while $r = 4.5\; \rm in$ . Therefore, when the lateral side of this cylinder is unrolled:

The width of the rectangle will be $6\; \rm in$ , while
The length of the rectangle will be $2 \, \pi \times 4.5\; \rm in = 9\, \pi\; \rm in$ .

That corresponds to a lateral surface area of $54\, \pi\; \rm in^2$ .

For cylinder $\rm Y$ , $h = 10.5\; \rm in$ while $r = 3\; \rm in$ . Similarly, when the lateral side of this cylinder is unrolled:

The width of the rectangle will be $10.5\; \rm in$ , while
The length of the rectangle will be $2\pi\times 3\; \rm in = 6\,\pi \; \rm in$ .

That corresponds to a lateral surface area of $63\,\pi \; \rm in^2$ .

Therefore, the lateral surface area of cylinder $\rm X$ is smaller than that of cylinder $\rm Y$ by $9\,\pi\; \rm in^2$ .

5 0

3 years ago