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tester [92]
3 years ago
11

I need help with number 7, and the second part of number 8. Thank you!

Mathematics
1 answer:
tangare [24]3 years ago
5 0

Answer:

7. \dfrac{\pi }{3},\ \pi ,\ \dfrac{5\pi }{3}

8. b. 0,\ \dfrac{2\pi }{3},\ \pi,\ \dfrac{4\pi }{3}

Step-by-step explanation:

7. Solve the equation

\cos x+\cos 2x=0

First, note that

\cos 2x=2\cos^2x-1

Hence,

\cos x+2\cos ^2x-1=0

Use substitution

t=\cos x,

then

2t^2+t-1=0\\ \\D=1^2-4\cdot 2\cdot (-1)=1+8=9\\ \\\sqrt{D}=\sqrt{9}=3\\ \\t_{1,2}=\dfrac{-1\pm 3}{2\cdot 2}=-1,\ \dfrac{1}{2}

Now,

\cos x=-1\ \text{or}\ \cos x=\dfrac{1}{2}

Solve each equation separately for 0\le x

\cos x=-1\\ \\x=\pi+2\pi k, \in Z\\ \\\text{only } x=\pi \in [0,2\pi)

\cos x=\dfrac{1}{2}\\ \\x=\pm\arccos \dfrac{1}{2}+2\pi k,\ k\in Z\\ \\x=\dfrac{\pi }{3}\ \text{and}\ x=\dfrac{5\pi }{3}

8. b. Solve the equation

\sin x+2\sin x\cos x=0

Rewrite it:

\sin x(1+2\cos x)=0

By zero product property,

\sin x=0\ \text{or}\ 1+2\cos x=0

Solve each equation separately.

\sin x=0\\ \\x=\pi k,\ k\in Z\\ \\\text{for}\ x\in [0,2\pi),\ \ \ \ x_1=0\ \ x_2=\pi

1+2\cos x=0\\ \\\cos x=-\dfrac{1}{2}\\ \\x=\pm \arccos \left(- \dfrac{1}{2}\right)+2\pi k,\ k\in Z\\ \\x=\dfrac{2\pi }{3}\ \text{and}\ x=\dfrac{4\pi }{3}

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Step-by-step explanation:

\vec{u}  =   (18,- 7)\Rightarrow  \vec{v}  = ( - 7,18) \Rightarrow  3\vec{v}  = ( - 21,54)

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It will take Mr. warner 15 1/2 hours to install a fence. Each day he works 1 1/3 hours on the fence. After he finished working o
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15.5-3.5=12. this is how many hours he has worked already.

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Factor 4x^2 -6w+2x-12xw
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Read 2 more answers
−11b+7=40<br> Like what am I supposed to do??
Nataly [62]

Answer:

The answer is 3

Step-by-step explanation:

First write the equation as you see it.

-11b+7=40

Then switch the take away seven from the left side then place it in the right side but as a minus.

-11b=40-7

Next do the math on the right side.

-11b=33

Once you have done that you divide the number on right with the number on the left which leaves you with the answer.

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Hope this helped :)

8 0
3 years ago
Use y = 16x^2 to complete the table.<br> What pattern (symmetry) do you notice from the table?
KATRIN_1 [288]

Answer:

In the table, we can see values of x and y, to complete it, we just need to input the different values of x in the equation:

y = 16*x^2

For example in the first slot, we have x = -4

Then we will have:

y = 16*(-4)^2 = 240

Then we complete the empty slot with 240.

The table would be:

\left[\begin{array}{cccccccccccccccc}x&-4&-3&-2.5&-2&-1.5&-1&-0.5&0&0.5&1&1.5&2&2.5&3&4\\y&240&\end{array}\right]

Now we just need to do the same for the other values:

y(-3) = 16*(-3)^2 = 144

y(-2.5) = 16*(-2.5)^2 = 100

y(-2) = 16*(-2)^2 = 64

y(-1.5) = 16*(-1.5)^2 = 36

y(-1) = 16*(-1)^2 = 16

y(-0.5) = 16*(-0.5)^2 = 4

y(0) = 16*(0)^2 = 0

y(0.5) = 16*(0.5)^2 = 4

y(1) = 16*(1)^2 = 16

y(1.5) = 16*(1.5)^2 = 36

y(2) = 16*(2)^2 = 64

y(2.5) = 16*(2.5)^2 = 100

y(3) = 16*(3)^2 = 144

y(4) = 16*(4)^2 = 240

Then the complete table is:

\left[\begin{array}{cccccccccccccccc}x&-4&-3&-2.5&-2&-1.5&-1&-0.5&0&0.5&1&1.5&2&2.5&3&4\\y&240&144&100&64&36&16&4&0&4&16&36&64&10&144&240\end{array}\right]

Then we can see that there is a symmetry around the value x = 0.

7 0
3 years ago
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