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4 years ago

(20 points & brainliest)

Mathematics

1 answer:

Levart [38]4 years ago

5 0

I believe it’s 1 because anything set to the 0 power is automatically 1. I’m sorry if I’m wrong

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Please I need help. It’s due soon

DENIUS [597]

Answer:

the answer is odd

Step-by-step explanation:

4 0

3 years ago

Please solve with explanation (high points)

IgorLugansk [536]

So, we know that:

$A=\dfrac{1}{2}ab\\P=a+b+c\\a^2+b^2=c^2$

We also know that $(a+b)^2=a^2+2ab+b^2$ . We will use that fact to find $c$ in terms of $A$ and $P$ .

$A=\dfrac{1}{2}ab\Rightarrow 2ab=4A\\P=a+b+c\Rightarrow a+b=P-c$

Therefore:

$(P-c)^2=c^2+4A\\P^2-2cP+c^2=c^2+4a\\2cP=P^2-4A\\c=\dfrac{P^2-4A}{2P}$

7 0

2 years ago

3.15x4.12=<br><img src="https://tex.z-dn.net/?f=3.15x4.12%20%3D%20" id="TexFormula1" title="3.15x4.12 = " alt="3.15x4.12 = " ali

gulaghasi [49]

Answer:

1/61.8 its right valu of x

Step-by-step explanation:

here we find the first valu and second is also same but there isx so cross the eqal side and get same answer

4 0

3 years ago

simplify 3/18 ??????????????????????????????????????????????????????????????????????????????????????????????????????????????????

Anna [14]

The simplified version of 3/18 is 1/6

7 0

3 years ago

What values of b satisfy 3(2b + 3)2 = 36?

KiRa [710]

the correct question is

What values of b satisfy 3(2b+3)^2 = 36

we have

$3(2b+3)^2 = 36$

Divide both sides by $3$

$(2b+3)^2 = 12$

take the square root of both sides

$( 2b+3)} =(+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3$

$b1=\frac{\sqrt{12}}{2} -\frac{3}{2}$

$b1=\sqrt{3} -\frac{3}{2}$

$b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}$

$b2=-\sqrt{3} -\frac{3}{2}$

therefore

the answer is

the values of b are

$b1=\sqrt{3} -\frac{3}{2}$

$b2=-\sqrt{3} -\frac{3}{2}$

6 0

3 years ago

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