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3 years ago

12

WZ = 32, YZ = 6, and X is the midpoint of WY

1 answer:

ZanzabumX [31]3 years ago

8 0

Answer:

$XY=13\ units$

Step-by-step explanation:

<u><em>The question is</em></u>

Find the value of XY

see the attached figure to better understand the problem

we know that

$WZ=WX+XY+YZ$ -----> by segment addition postulate

If X is the midpoint WY

then

$WX=XY$

substitute in the formula above

$WZ=XY+XY+YZ$

$WZ=2XY+YZ$

substitute the given values and solve for XY

$32=2XY+6$

Subtract 6 both sides

$32-6=2XY$

$26=2XY$

Divide by 2 both sides

$26/2=XY$

Rewrite

$XY=13\ units$

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3 years ago

The Smith Family is designing new plans for an in ground pool. Mr. Smith draws a rectangular shape with a length that is 5 feet

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To solve this problem you must apply the proccedure shown below:

1- The formula for calculate the area of this rectangle is:

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3. If the length is $5ft$ longer than the heigth, you have:

$L=x+5$

4. Therefore, you can write the following polynomial expression:

$A(x)=x(x+5)\\ A(x)=x^{2} +5x$

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3 years ago

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3 years ago

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3 years ago

Find the following quotients, and write the quotient in standard form. (a)- x^2-9 ÷ x-3 (b)- x^3-27 ÷ x-3 (c) x^4- 81 ÷ x-3

IgorLugansk [536]

Answer:

(a) x+3

(b) $x^2+3x+9$

(c) $(x+3)(x^2+9)$

Step-by-step explanation:

We have given that

(a) $\frac{x^2-9}{x-3}$

From the algebraic identity we know that

$a^2-b^2=(a+b)(a-b)$

So $\frac{x^2-9}{x-3}=\frac{(x+3)(x-3)}{x-3}=x+3$

(b) $\frac{x^3-27}{x-3}$

We know the algebraic identity

$a^3-b^3=(a-b)(a^2+ab+b^2)$

So $\frac{x^3-27}{x-3}=\frac{x^3-3^3}{x-3}=\frac{(x-3)(x^2+3x+9)}{x-3}=x^2+3x+9$

(c) We have given $\frac{x^4-81}{x-3}$

We know the algebraic identity

$a^2-b^2=(a+b)(a-b)$

$\frac{x^4-81}{x-3}=\frac{(x^2)^2-(3^2)^2}{x-3}=\frac{(x^2-9)(x^2+9)}{x-3}=\frac{(x+3)(x-3)(x^2+9)}{x-3}=(x+3)(x^2+9)$

3 0

3 years ago