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LekaFEV [45]
3 years ago
8

Find values of 3sin^2x-2cosx-3=0

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
4 0

Answer:

Step-by-step explanation:

3 sin²x-2cos x-3=0

3(1-cos² x)-2 cos x-3=0

3-3 cos²x-2 cos x-3=0

-3 cos ²x-2 cos x=0

cos x( -3cos x-2)=0

cos x=0

x=(2n+1)π/2,where n is an integer.

or -3 cos x-2=0

3 cos x=-2

cos x=-2/3

x=cos ^{-1}(-2/3)≈131.81°

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PLEASE HELP ME OUT HERE!!!!<br> THIS IS DUE TODAY!!!!!
dolphi86 [110]

Answer:

I'm pretty sure the answer is b

5 0
3 years ago
Read 2 more answers
. What is the negative solution to this equation?<br> 8x2 – 4x = 363
stiv31 [10]

Answer:

- x = 81 and 3/4

Step-by-step explanation:

8 × 2 – 4x = 363

16 - 4x = 363

Subtract 16 from each side:

16 - 16 - 4x = 363 - 16

- 4x = 327

Divide each side by 4:

- 4x  ÷ 4 = 327 ÷ 4

- x = 81 and 3/4

8 0
3 years ago
WILL MARK BRAINLIEST!! The coordinates of the vertices of quadrilateral JKLM are J(-3,2), K(4,-1), L(2,-5) and M(-5,-2). Find th
irinina [24]
Answer:
JKLM is a parallelogram
Explanation:
The slope m of a line through two points (x1,y1) and (x2,y2)is given by the formula:
m=Δy/Δx=y2-y1/x2-x1
So the slopes of the sides of our quadrilateral are:
mJK=(−1)−24−(−3)=−37
mKL=(−5)−(−1)2−4=2
mLM=(−2)−(−5)−5−2=−37
mMJ=2−(−2)(−3)−(−5)=2
So JK is parallel to LM and KL is parallel to MJ
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5 0
3 years ago
The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest
Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)

Calculation the value from standard normal z table, we have,  

P(x < 5) =0.7287= 72.87\%

b) P( port handles 3 or more million tons of cargo per week)

P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)

Calculating the value from the standard normal table we have,

1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%

c)P( port handles between 3 million and 4 million tons of cargo per week)

P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%

P(3 \leq x \leq 4) = 23.7\%

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85  

= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85  

=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15  

Calculation the value from standard normal z table, we have,  

P( z \leq -1.036) = 0.15

\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

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