HELP! WILL MARK BRAINLIEST WITH A EXPLANATION!!!!please & thank u

HELP FAST FAST FAST FAST PLS ILL LOVE YOU FORVER. Greta is putting a 20 foot fence around her pool. She uses all the border to m

dezoksy [38]

Answer:

20 over 4 llllllllll<em /><em />

8 0

2 years ago

What is an equation of the line that passes through the point (-6, – 7) and is perpendicular to the line 6x + 5y = 30?

fomenos

Answer :6x + 5y = 30

5y = 30 - 6x

y = 6 - (6/5) x

So slope = -6/5

Perpendicular line has slope 5/6 (product of the slopes is -1)

So y = 5/6 x + b

Passed through (-6, -7) so -7 = -5 + b

Sp b = -2

So y = (5/6)x - 2

Step-by-step explanation:

3 0

2 years ago

The answer is correct? If it’s wrong please explain me how to solve it :’)

cricket20 [7]

Answer:

48

Step-by-step explanation:

x = -2 y =3

3x² - 2xy²

3(-2)² -2(-2)(3)²

square first

3*4 -2(-2)*9

multiplty from left to right

12-2(-2)*9

multiply from left to right again

12 -(-36) subtract using rule for integers that double minus is plus

12 + 36

48

7 0

2 years ago

Can someone helpppp

Dmitry [639]

Yes, I can help what would you like me to do

5 0

2 years ago

If cos theta= -8/17 and theta is in quadrant 3, what is cos2 theta and tan2 theta

Karo-lina-s [1.5K]

$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad 
\begin{array}{llll}
\textit{now, hypotenuse is always positive}\\
\textit{since it's just the radius}
\end{array}
\\\\\\
thus\qquad cos(\theta)=\cfrac{-8}{17}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}$

since the hypotenuse is just the radius unit, is never negative, so the - in front of 8/17 is likely the numerator's, or the adjacent's side

now, let us use the pythagorean theorem, to find the opposite side, or "b"

$\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite
\end{cases}
\\\\\\
\pm\sqrt{17^2-(-8)^2}=b\implies \pm\sqrt{225}=b\implies \pm 15=b$

so... which is it then? +15 or -15? since the root gives us both, well
angle θ, we know is on the 3rd quadrant, on the 3rd quadrant, both, the adjacent(x) and the opposite(y) sides are negative, that means, -15 = b

so, now we know, a = -8, b = -15, and c = 17
let us plug those fellows in the double-angle identities then

$\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
\boxed{1-2sin^2(\theta)}\\
2cos^2(\theta)-1
\end{cases}
\\ \quad \\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
-----------------------------\\\\
cos(2\theta)=1-2sin^2(\theta)\implies cos(2\theta)=1-2\left( \cfrac{-15}{17} \right)^2
\\\\\\
cos(2\theta)=1-\cfrac{450}{289}\implies cos(2\theta)=-\cfrac{161}{289}$

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\left( \frac{-15}{-8} \right)}{1-\left( \frac{-15}{-8} \right)^2}
\\\\\\
tan(2\theta)=\cfrac{\frac{15}{4}}{1-\frac{225}{64}}\implies tan(2\theta)=\cfrac{\frac{15}{4}}{-\frac{161}{64}}
\\\\\\
tan(2\theta)=\cfrac{15}{4}\cdot \cfrac{-64}{161}\implies tan(2\theta)=-\cfrac{240}{161}$

6 0

2 years ago