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Ronch [10]
3 years ago
9

QUICKK

Mathematics
1 answer:
VLD [36.1K]3 years ago
7 0
3x + 4(2) = 20
3x + 8 = 20
3x = 12
x = 4
(4,2)
You have y=2 then all you need to do is plug 2 into the y in the equation 3x + 4y = 20 and do the math to find x. Hope this helps
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Please help me thank you
Hoochie [10]

Answer:

\large\boxed{\sin2\theta=\dfrac{\sqrt3}{2},\ \cos2\theta=\dfrac{1}{2}}

Step-by-step explanation:

We know:

\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet

We have

\sin\theta=\dfrac{1}{2}

Use \sin^2\theta+\cos^2\theta=1

\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}

\sin2\theta=2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}\\\\\cos2\theta=\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}

3 0
3 years ago
Y=3x+18 <br> -12x+4y=-24 linear system of equations
solong [7]

Answer:

Step-by-step explanation:

Use substitution, y=3x+18 so -12x+4(3x+18)=-24. From here you can try to solve for x

-12x+12x+72=-24

The xs cancel, which leaves you with 72=-24

Since 72 cannot equal -24 there is no answer to this system of equations

7 0
3 years ago
Solve this question to get 11 points! <br> 6(x + 3) = 21<br> x=?
azamat
6 (x + 3) = 21
X= 3
4 0
3 years ago
Solve for a. 15(25−5a)=4−a
VMariaS [17]
15(25-5a)=4-a
375-75a=4-a
Add 75a to both sides
375= 4+74a
Subtract 4 from both sides
371=74a
Divide both by 74
5.01=a
6 0
3 years ago
Read 2 more answers
Helppppppppppppp plssssssssssss
oksian1 [2.3K]

Pretty sure the first one is C

3 0
3 years ago
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