We know that two complements add up to 90.
Let's call the smaller angle x and the larger y.
3x = y
x + y = 90
We can use simple substitution.
x + 3x = 90
4x = 90
x = 22.5

Then, since we know that 3x=y, we can find the larger angle.
3*22.5 = 67.5

4 0

3 years ago

-4rover4+5rover3 please help me

Alchen [17]

$Solution, \frac{-4r}{4}+\frac{5r}{3}=\frac{2r}{3}$

$Steps:$

$\frac{-4r}{4}+\frac{5r}{3}$

$\frac{-4r}{4},\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b},\\-\frac{4r}{4},\\\mathrm{Divide\:the\:numbers:}\:\frac{4}{4}=1,\\-r,\\-r+\frac{5r}{3}$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:r=\frac{r3}{3},\\\frac{5r}{3}-\frac{r\cdot \:3}{3}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c},\\\frac{5r-r\cdot \:3}{3}$

$5r-r\cdot \:3,\\\mathrm{Add\:similar\:elements:}\:5r-3r=2r,\\\frac{2r}{3}$

$The\:Correct\:Answer\:Is\:\frac{2r}{3}$

$Hope\:This\:Helps!!!$

$-Austint1414$

5 0

3 years ago

(3x+4)^2=14 what are the solutions?

kherson [118]

$\bf (3x+4)^2=14\implies \sqrt{(3x+4)^2}=\pm\sqrt{14}\implies 3x+4=\pm\sqrt{14}
\\\\\\
3x=\pm \sqrt{14}-4\implies x=\cfrac{\pm \sqrt{14}-4}{3}\implies x=
\begin{cases}
\cfrac{\sqrt{14}-4}{3}\\\\
\cfrac{- \sqrt{14}-4}{3}
\end{cases}$

7 0

3 years ago