$\bf c^2=a^2+b^2\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
------\\
a=m^2-n^2\\
b=2mn
\end{cases}\\\\\\ c^2=(m^2-n^2)^2+(\underline{2mn})^2
\\\\\\
c^2=m^4-2m^2n^2+n^4+\underline{2^2m^2n^2}
\\\\\\
c^2=m^4-2m^2n^2+n^4+{4m^2n^2}
\\\\\\
c^2=m^4+2m^2n^2+n^4\impliedby \textit{perfect square trinomial}
\\\\\\
c^2=(m^2+n^2)^2\implies c=\sqrt{(m^2+n^2)^2}\implies c=m^2+n^2$

7 0

3 years ago

Log8(216)+[log(42)_log(6)]÷log(49)

makkiz [27]

Answer:

3 log (3) + 1/2

Step-by-step explanation:

log8(216) + [log(42) - log(6)] ÷ log(49) =
log (216)/ log (8) + [log (7) + log (6) - log (6)] ÷ 2 log (7) =
3 log (3) log (2) / 3 log (2) + log (7) / 2 log (7) =
3 log (3) + 1/2

7 0

3 years ago

I WILL GIVE BRAINLY

Simora [160]

Answer:

Step-by-step explanation:

We will let d represent the amount of dimes Angela has.

Angel has 5 fewer quarters than twice the number of dimes.

Twice the number of dimes will be 2d.

Then, 5 fewer will yield 2d-5.

Therefore, the amount of quarters Angela has is 2d-5.

Then, the total amount of coins (dimes and quarters) together, will be:

$d+(2d-5)$

The first part is the amount of dimes.

And the second part is, as we determined, the amount of quarters.

Therefore, our answer is D.

7 0

2 years ago