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ipn [44]
3 years ago
10

Directions : Factor each of the following Differences of two squares and write your answer together with solution​

Mathematics
2 answers:
N76 [4]3 years ago
4 0

\huge \boxed{\mathfrak{Question} \downarrow}

Factor each of the following differences of two squares and write your answer together with solution.

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

<h3><u>1. x² - 36</u></h3>

\sf \: x ^ { 2 } - 36

Rewrite \sf\:x^{2}-36 as x^{2}-6^{2}. The difference of squares can be factored using the rule:\sf\: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

\boxed{ \boxed{ \bf\left(x-6\right)\left(x+6\right) }}

__________________

<h3><u>2. 49 - x²</u></h3>

\sf \: 49 - x ^ { 2 }

Rewrite 49-x² as 7²-x². The difference of squares can be factored using the rule: \sf\:a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

\sf \: \left(7-x\right)\left(7+x\right)

Reorder the terms.

\boxed{ \boxed{ \bf\left(-x+7\right)\left(x+7\right) }}

__________________

<h3><u>3. 81 - c²</u></h3>

\sf \: 81 - c ^ { 2 }

Rewrite 81-c²as 9²-c². The difference of squares can be factored using the rule: \sf\:a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

\sf\left(9-c\right)\left(9+c\right)

Reorder the terms.

\boxed{ \boxed{ \bf\left(-c+9\right)\left(c+9\right) }}

__________________

<h3><u>4</u><u>.</u><u> </u><u>m²</u><u>n</u><u>²</u><u> </u><u>-</u><u> </u><u>1</u></h3>

\sf \: m ^ { 2 } n ^ { 2 } - 1

Rewrite m²n² - 1 as \sf\left(mn\right)^{2}-1^{2}. The difference of squares can be factored using the rule: \sf\:a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

\boxed{ \boxed{ \bf\left(mn-1\right)\left(mn+1\right) }}

OverLord2011 [107]3 years ago
3 0

Step-by-step explanation:

please mark me as brainlest

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A 4 metre ladder is placed against a vertical wall.
Black_prince [1.1K]

Answer:

Original position: base is 1.5 meters away from the wall and the vertical distance from the top end to the ground let it be y and length of the ladder be L.

Step-by-step explanation:

By pythagorean theorem, L^2=y^2+(1.5)^2=y^2+2.25 Eq1.

Final position: base is 2 meters away, and the vertical distance from top end to the ground is y - 0.25 because it falls down the wall 0.25 meters and length of the ladder is also L.

By pythagorean theorem, L^2=(y -0.25)^2+(2)^2=y^2–0.5y+ 0.0625+4=y^2–0.5y+4.0625 Eq 2.

Equating both Eq 1 and Eq 2: y^2+2.25=y^2–0.5y+4.0625

y^2-y^2+0.5y+2.25–4.0625=0

0.5y- 1.8125=0

0.5y=1.8125

y=1.8125/0.5= 3.625

Using Eq 1: L^2=(3.625)^2+2.25=15.390625, L=(15.390625)^1/2= 3.92 meters length of ladder

Using Eq 2: L^2=(3.625)^2–0.5(3.625)+4.0625

L^2=13.140625–0.90625+4.0615=15.390625

L= (15.390625)^1/2= 3.92 meters length of ladder

<em>hope it helps...</em>

<em>correct me if I'm wrong...</em>

4 0
3 years ago
Leave the answer as an improper fraction​
Ghella [55]

Hey there!

ANSWER: \frac{23}{20}

EXPLANATION:

To find the answer to your question, you will need to add.

\frac{3}{4}+\frac{2}{5}

First, multiply the denominator.

5*4=20

So the denominator will be 20.

Now move on to the numerator. Multiply 5 by 3.

5*3=15

Now multiply 4 bu 2.

4*2=8

If you want to get the numerator, add what you got after multiply.

15+8=23

The numerator is 23. So now let's complete the fraction.

\frac{23}{20}

This is your answer!

Hope this helps!

-TestedHyperr

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Answer:

81

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