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3 years ago

10

Please answer this question:)

2 answers:

Licemer1 [7]3 years ago

7 0

Answer:

Social interaction and religion

Step-by-step explanation:

The first settlers move due to the lack of freedom of religeon in Britain.

olga_2 [115]3 years ago

5 0

Answer:

a or the first one

Step-by-step explanation:

built:Different

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Find the area of the following composite figure

just olya [345]

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1713

Step-by-step explanation:

952+288+472=1713

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3 years ago

Simplify this expression: -2x + 3 – (5 – 6x).

Fittoniya [83]

Step-by-step explanation:

-2x + 3 – (5 – 6x)

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4 years ago

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-25^-1/2 with explanation please

slavikrds [6]

Answer:

The simplified value of $-25^{-\frac{1}{2}}$ is $-\frac{1}{5}$ .

Step-by-step explanation:

As the given expression is

$-25^{-\frac{1}{2}}$

Lets simplify this expression step by step

As

$-25^{-\frac{1}{2}}....[A]$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$25^{-\frac{1}{2}}=\frac{1}{25^{\frac{1}{2}}}$

So, lets plug $25^{-\frac{1}{2}}=\frac{1}{25^{\frac{1}{2}}}$ in equation [A] i.e. $-25^{-\frac{1}{2}}....[A]$

$-25^{-\frac{1}{2}}....[A]$

$=-\frac{1}{25^{\frac{1}{2}}}$ ∵ $25^{-\frac{1}{2}}=\frac{1}{25^{\frac{1}{2}}}$

$=-\frac{1}{5}$ ∵ $25^{\frac{1}{2}}=5$

Therefore, the simplified value of $-25^{-\frac{1}{2}}$ is $-\frac{1}{5}$ .

Keywords: simplification, exponent rule

Learn more about expression simplification from brainly.com/question/13848834

#learnwithBrainly

5 0

3 years ago

I really need help guys can some one plz help me it's for 10 points

11111nata11111 [884]

Answer:

$\sqrt{8}----- 2units,2units$

$\sqrt{7} ------\sqrt{5}units,\sqrt{2}units$

$\sqrt{5}-------1unit,2units$

$3------2units, \sqrt{5}units$

Step-by-step explanation:

Use pythagorean's theorem to solve each pair individually.

The instructions say that you have the two sides (a and b) and you have to match it with their hypothenuse.

$Formula: c^2=a^2+b^2$

1. $a=\sqrt{5}, b=\sqrt{2}$

Plug this into the formula.

$c=\sqrt{(\sqrt{5} )^2+(\sqrt{2})^2 }$

The square here eliminates the roots.

$c=\sqrt{5+2}\\ c=\sqrt{7}$

2. $a=\sqrt{3}, b=4$

$c=\sqrt{(\sqrt{3} )^2+(4)^2}\\ c=\sqrt{3+16}\\ c=\sqrt{19}$

3. $a=2, b=\sqrt{5}$

$c=\sqrt{(2)^2+(\sqrt{5})^2 }\\ c=\sqrt{4+5}\\ c=\sqrt{9}\\ c=3$

4. $a=2, b=2$

$c=\sqrt{(2)^2+(2)^2}\\ c=\sqrt{4+4}\\ c=\sqrt{8}$

5. $a=1, b=2$

$c=\sqrt{(1)^2+(2)^2}\\ c=\sqrt{1+4}\\ c=\sqrt{5}$

7 0

3 years ago