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hjlf
3 years ago
15

A company is creating a box without a top from a piece of cardboard, but cutting out square corners with side length x.

Mathematics
2 answers:
Volgvan3 years ago
6 0
Answer: Choice B) The expression (10-2x)(30-2x)x represents the volume of the box

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The original width is 10 inches. The width reduces to 10-2x inches after we cut off the top two corners of the rectangle. We can think of it as taking 10 and subtracting off two copies of x like so: 10-x-x = 10-2x

Similarly, the length goes from 30 inches to 30-2x inches. This time we're taking off the top and bottom corners (focus on either side it doesn't matter). 

The height of the box is x inches due to this portion being folded up. 

Volume of box = (width)*(length)*(height)
Volume of box = (10-2x)(30-2x)x

Note: the units for the answer are in cubic inches which can be written as "in^3" (inches cubed). 

pantera1 [17]3 years ago
6 0

Answer:

(10−2x)(30−2x)x

Step-by-step explanation:

I know how to explain it, but the other person's answer already has a good explanation. I'm just confirming this to be correct! :D

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Aleonysh [2.5K]

The location of the y value of R' after using the translation rule is -10

<h3>What will be the location of the y value of R' after using the translation rule? </h3>

The translation rule is given as:

(x + 4, y - 7)

The pre-image of R is located at (-17, -3)

Rewrite as

R = (-17, -3)

When the translation rule is applied, we have:

R' = (-17 + 4, -3 - 7)

Evaluate

R' = (-13, -10)

Remove the x coordinate

R'y = -10

Hence, the location of the y value of R' after using the translation rule is -10

Read more about translation at:

brainly.com/question/26238840

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5 0
1 year ago
Which expression is equivalent to square root 55x^7y6/11x^11y^8
Novay_Z [31]

Answer:

\large\boxed{\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\dfrac{\sqrt5}{x^2y}}

Step-by-step explanation:

\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{55}{11}\cdot\dfrac{x^7}{x^{11}}\cdot\dfrac{y^6}{y^8}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=\sqrt{5x^{7-11}y^{6-8}}=\sqrt{5x^{-4}y^{-2}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt5\cdot\sqrt{x^{-4}}\cdot\sqrt{y^{-2}}=\sqrt5\cdot\sqrt{x^{(-2)(2)}}\cdot\sqrt{y^{(-1)(2)}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt5\cdot\sqrt{(x^{-2})^2}\cdot\sqrt{(y^{-1})^2}\qquad\text{use}\ \sqrt{a^2}=a

=\sqrt5\cdot x^{-2}\cdot y^{-1}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\to a^{-1}=\dfrac{1}{a}\\\\=\sqrt5\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}=\dfrac{\sqrt5}{x^2y}

6 0
3 years ago
Use the Distributive Property to rewrite
Stella [2.4K]

Answer:

c

Step-by-step explanation:

6 0
2 years ago
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