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3 years ago

Twenty-one is 20% of ________.

Mathematics

1 answer:

Elanso [62]3 years ago

4 0

To find the answer for this, divide 21 by the percentage:

21 / 20% = 21 / 0.20 = 105

21 is 20% of 105

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Write the polynomial as a square of a binomial or as an expression opposite to a square of a binomial:

spayn [35]

Answer:

A) $0.25x^2-0.6xy+0.36y^2=\left(0.5x-0.6y\right)^2$

B) $-a^2+0.6a-0.09=-\left(10a-3\right)^2$

C) $\frac{9a^4}{16}+a^3+\frac{4a^2}{9}=a^2(\left(9a+8\right)^2)$

D) $-16m^2-24mn -9n^2=-\left(4m+3n\right)^2$

Step-by-step explanation:

The square of a binomial is the sum of: the square of the first terms, twice the product of the two terms, and the square of the last term.

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

To find the square of the binomial of the following polynomials you must:

A) $0.25x^2-0.6xy+0.36y^2$

Apply radical rule: $a=\left(\sqrt{a}\right)^2$

$0.25=\left(\sqrt{0.25}\right)^2\\0.36=\left(\sqrt{0.36}\right)^2$

$\left(\sqrt{0.25}\right)^2x^2-0.6xy+\left(\sqrt{0.36}\right)^2y^2$

Apply exponent rule: $a^mb^m=\left(ab\right)^m$

$\left(\sqrt{0.25}\right)^2x^2=\left(\sqrt{0.25}x\right)^2\\\left(\sqrt{0.36}\right)^2y^2=\left(\sqrt{0.36}y\right)^2$

$\left(\sqrt{0.25}x\right)^2-0.6xy+\left(\sqrt{0.36}y\right)^2$

Rewrite $0.6xy$ as $2\cdot \:0.5x\cdot \:0.6y$

$\left(\sqrt{0.25}x\right)^2-2\cdot \:0.5x\cdot \:0.6y+\left(\sqrt{0.36}y\right)^2$

Apply perfect square formula: $\left(a-b\right)^2=a^2-2ab+b^2$

$a=0.5x,\:b=0.6y$

$\left(\sqrt{0.25}x\right)^2-2\cdot \:0.5x\cdot \:0.6y+\left(\sqrt{0.36}y\right)^2=\left(0.5x-0.6y\right)^2$

B) $-a^2+0.6a-0.09$

Multiply both sides by 100

$-a^2\cdot \:100+0.6a\cdot \:100-0.09\cdot \:100\\-100a^2+60a-9$

Factor out common term -1

$-\left(100a^2-60a+9\right)$

Break the expression into groups and factor out common terms

$-(\left(100a^2-30a\right)+\left(-30a+9\right))\\-(10a\left(10a-3\right)-3\left(10a-3\right))\\-(\left(10a-3\right)\left(10a-3\right))\\-\left(10a-3\right)^2$

C) $\frac{9a^4}{16}+a^3+\frac{4a^2}{9}$

Apply exponent rule: $a^{b+c}=a^ba^c$

$a^3=aa^2\\a^4=a^2a^2$

$\frac{9a^2a^2}{16}+aa^2+\frac{4a^2}{9}$

Factor out common term $a^2$

$a^2\left(\frac{9a^2}{16}+a+\frac{4}{9}\right)$

Factor $\frac{9a^2}{16}+a+\frac{4}{9}\right$

Find the Least Common Multiplier (LCM) of 16, 9 which is 144.

Multiply by LCM

$\frac{9a^2}{16}\cdot \:144+a\cdot \:144+\frac{4}{9}\cdot \:144\\81a^2+144a+64$

$81a^2+144a+64=\left(9a\right)^2+2\cdot \:9a\cdot \:8+8^2$

Apply perfect square formula: $\left(a+b\right)^2=a^2+2ab+b^2$

$a=9a,\:b=8$

$81a^2+144a+64=\left(9a+8\right)^2$

$\frac{9a^4}{16}+a^3+\frac{4a^2}{9}=a^2(\left(9a+8\right)^2)$

D) $-16m^2-24mn -9n^2$

Factor out common term -1

$-\left(16m^2+24mn+9n^2\right)$

Break the expression into groups and factor out common terms

$\left(16m^2+12mn\right)+\left(12mn+9n^2\right)\\4m\left(4m+3n\right)+3n\left(4m+3n\right)\\\left(4m+3n\right)\left(4m+3n\right)\\-\left(4m+3n\right)\left(4m+3n\right)\\-\left(4m+3n\right)^2$

$-16m^2-24mn -9n^2=-\left(4m+3n\right)^2$

3 0

3 years ago

This is nothing you need to solve, I just need a refresher about this type of math. For some reason I can't remember this right

SVEN [57.7K]

Answer:

Yes there is (but we like to remove it)

Step-by-step explanation:

Recall that x^1 = x, which represents x being multiplied by itself once, so you just have x. So, we usually take out the exponent 1 because it is not necessary since it's just the same as x.

6 0

2 years ago

Read 2 more answers

A drawing of a room has a scale of 1 inch = 4 feet . If the actual dimensions of the room are 14 feet by 16 feet , what are the

hammer [34]

Answer:

3.2(roughly) inches by 4 inches

Step-by-step explanation:

7 0

4 years ago

HELPPP PLEASEEEE

Varvara68 [4.7K]

Najsjssjjsjsjsjsjdisisiss

8 0

3 years ago

Using the slope formula, find the slope of the line through the given points. (7,4) and (5,6)

anyanavicka [17]

We can use the points (7, 4) and (5, 6) to solve.

Slope formula: y2-y1/x2-x1

= 6-4/5-7

= 2/-2

= -1

______

Best of Luck,

Wolfyy :)

4 0

3 years ago