I need help with this problem

Mathematics

2 answers:

Setler79 [48]3 years ago

8 0

One angle is 40 degrees and the other is 50 degrees.

A complementary angle is 90 degrees.

As the ratio is 4:5 we can use the equation: 4x+5x=90

Solving it we get:
9x=90
x=10.

So the ratio is 4(10):5(10)

Meaning the two angles are 40 degrees and 50 degrees.

I hope this helps!
(And if you can, could I have brainliest please?)

Sergio039 [100]3 years ago

6 0

The answer is 57765363535366355336

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Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)

Leni [432]

Alrighty

remember some rules
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}=\frac{1}{x^m}$
and
$x^0=1$ for all real values of x
and
$(a^b)^c=a^{bc}$
and
$(\frac{a}{b})^c=\frac{a^c}{b^c}$
and
(a/b)/(c/d)=(ad)/(bc)
and
$(a^b)(a^c)=a^{b+c}$
and
$\frac{a^m}{a^n}=a^{m-n}$
and don't forget pemdas
example: $-x^m=-1(x^m)$ but $(-x)^m=(-1)^m(x^m)$

so

4.
$(\frac{c^{-2}}{2})^{-2}=$

$\frac{(c^{-2})^{-2}}{2^{-2}}=$

$\frac{c^4}{\frac{1}{2^2}}=$

$\frac{c^4}{\frac{1}{4}}=$

$4c^4$

5.

$\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=$

$\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=$

$\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=$

$\frac{(a^4bc^5)b^3}{(-1)(a^2)}=$

$\frac{-a^4b^4c^5}{a^2}=$

$-a^2b^4c^5$