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3 years ago

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Grace wove a potholder with an area of 80 square inches. The lengths and widths of the sides are whole numbers. Which dimensions

make the most sense for a potholder?

1 answer:

docker41 [41]3 years ago

7 0

Answer:

Kindly check explanation

Step-by-step explanation:

Given the question :

Grace wove a potholder with an area of 80 square inches. The lengths and widths of the sides are whole numbers. Which dimensions make the most sense for a potholder?

Since the Area of the potholder = 80 sq inch

And the dimension of the potholder ; length and width are integers ; Hence, possible dimensions could be ;

Area = length × width

80 = length × width

(80, 1), (40, 2), (20, 4), (10, 8), (16, 5)

Potholders are fabrics sewn from the purpose of being used to handle pots and other kitchen equipments.

Usually the dimension are usually close, hence, the most sensible dimension a potholder with an area of 80 sq inch could have is 8 by 10 or vice-versa.

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Try this B E Given: BELO, BOLE Prove ABEL ALOB Complete the Proof Statements 1. 2. BO=LE 3. 4. ∆BEL =∆LOB REASONS 1.GIVEN 2. 3.

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The given congruency of the sides $\mathbf{\overline{BE}}$ and $\mathbf{\overline{LO}}$ and $\mathbf{\overline{BO}}$ and $\mathbf{\overline{LE}}$ as well

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<h3>Which values correctly completes the table?</h3>

The completed two column proof is presented as follows;

Statement ${}$ Reasons

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2 years ago

Use slopes and y-intercepts to determine if the lines 5x-5y=-2 and -x+2y=4

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The given pair of lines are not perpendicular.

<h3>What is a line?</h3>

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1 year ago

The area of a rectangle is 440 square inches. The length of the rectangle is 29 inches more than its width. What is the length o

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3 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

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4 years ago

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