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vova2212 [387]
2 years ago
14

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal plac

es where appropriate. If there is no solution, enter NO SOLUTION.)
A). (tan(x) − 4)(9 sin^2(x) − 1) = 0

B). 3 sin(2x) − 4 sin(x) = 0
Mathematics
1 answer:
Serhud [2]2 years ago
3 0

Answer:

Step-by-step explanation:

A)(tanx-4)(9sin^2x-1)=0\\\\\\

9sin^2x-1=0\\\\sin^2x=\frac{1}{9} sinx=\frac{1}{3} or sinx=\frac{-1}{3}

OR    tanx=4\\

B)

3sin2x=3.2.sinx.cosx\\\\6.sinx.cosx-4.sinx= 2sinx.(3cosx-2)=0\\\\3.cosx=2 cosx=\frac{2}{3}

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Write an equation for the quadratic graphed below
uranmaximum [27]

Answer:

Step-by-step explanation:

LOL The graph doesn’t match the y intercept :)

Anyway

If we have point (0,-3) we have a quadratic of

y=ax^2+bx-3 we are given points (-1,0) and (2,0) so

a-b-3=0 and 4a+2b-3=0

4a+2b-3+2(a-b-3)=0

4a+2b-3+2a-2b-6=0

6a-9=0

6a=9

a=1.5, since a-b=3

1.5-b=3

b=-1.5

y=1.5x^2-1.5x-3

6 0
3 years ago
Read 2 more answers
A kite was flying 50 ft high from ground level. Suddenly it loses 20 ft in height due to wind.
goldfiish [28.3K]

Answer:

30ft

Step-by-step explanation:

50- 20 = 30

hope it helps!

5 0
3 years ago
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Rounded to the nearest hundredth, what is the positive solution to the quadratic equation 0 = 2x2 + 3x – 8?
zmey [24]
2x² + 3x -8 = 0

a = 2 , b = 3 , c = -8
So..
D = b²-4ac = 9 + 64 = 73

for bigger solution , use x = -b + √D / 2a
For smaller solution, use x = -b - √D / 2a

So
= -3 + √73 / 2. 2
= 1/4 (√73 -3)

Hopefully helpfull.. ^_^
8 0
3 years ago
Read 2 more answers
WILL MARK THE BRAINLIEST
Anestetic [448]

Answer:

Step-by-step explanation:

65.97in

5 0
3 years ago
If 10% is deducted from a bill, $58.50 remains to be paid. How much is the original bill?
NARA [144]

Answer:

From the above we can roughly for an equation which is:

58.50 = Bill - (10% x Bill)

58.5 = Bill - (1/10 x Bill)

58.5 = 9/10 x Bill (Divide both sides by 9/10)

We get:

Bill = 58.5 / (9/10)

Bill = 65

Original bill is $65.

8 0
2 years ago
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