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jeka57 [31]
3 years ago
14

Consider the function y = cos (x + pi/6). Which of the following is true?

Mathematics
2 answers:
max2010maxim [7]3 years ago
7 0
The answer would be A. There is a phase shift to the left. Because you are adding pi/6.
kumpel [21]3 years ago
5 0

Answer with explanation:

To Answer this question,

→→We will look at the graph and observe the function carefully

     y = cos x

It has domain equal to [\frac{-\pi}{2}, \frac{-\pi}{2}]

and Range equal to , [-1, 1].

On , Y axis it takes maximum value =1 at , x=0 degree.

And→→ when you will look at the graph and observe the second  function,

 y=cos (x +\frac{\pi}{6})

Domain is equal to [\frac{-\pi}{2}+\frac{\pi}{6} ,\frac{\pi}{2}+\frac{\pi}{6}]=[\frac{-\pi}{3},\frac{2\pi}{3}]

Range =[-1, 1]

It's Maximum value is also =1, which it takes at , x=-30 degree.

And, when you will compare the graphs of two functions,the graph has been shifted

  \frac{\pi}{2} - \frac{\pi}{3}=\frac{\pi}{6}

to the left.

Option A: There is a phase shift to the left.

 

   

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Find the value of x.<br>A. 90 <br>B. 70 <br>C. 20 <br>D. 50​
emmasim [6.3K]

Answer: B. 70

Step-by-step explanation:

i actually can't tell but i think this is right

7 0
3 years ago
1.6E-8 is equal to what numbers or scientific notation?
prisoha [69]
It's equal to 1.6x10^(-8)
7 0
3 years ago
Find parametric equations for the line through the point (0, 2, 2) that is parallel to the plane x + y + z = 4 and perpendicular
Oxana [17]

Answer: X = 3t, Y =2 - t, Z =2

Step-by-step explanation: the plane

x + y + z =4has normal vector

M =<1,1,1> and the line

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v =<1, −1, 2>. So the vector

A= n × v

=<1, 1, 1> × <1, −1, 2>

=<2−(−1),1−2,−1−1>

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5 0
4 years ago
Given that α and β are the roots of the quadratic equation <img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B6x-7%3Dp" id="Tex
siniylev [52]

Answer:

\large \boxed{\sf \ \ \ p=-11 \ \ \ }

Step-by-step explanation:

Hello,

\alpha \text{ and } \beta \text{ are the roots of the following equation}

   2x^2+6x-7=p

It means that

   2\alpha^2+6\alpha-7=p \\\\2\beta ^2+6\beta -7=p \\\\

And we know that

\alpha= 2\cdot \beta

So we got two equations

   2(2\beta)^2+6\cdot 2 \cdot \beta -7=p \\\\8\beta^2+12\beta -7=p\\\\ and \ 2\beta ^2+6\beta -7=p \ So \\\\\\8\beta^2+12\beta -7 = 2\beta ^2+6\beta -7\\\\6\beta^2+6\beta =0\\\\\beta(\beta+1)=0\\\\ \beta =0 \ or \ \beta=-1

For \beta =0, \ \ \alpha =0, \ \ p = -7

For \beta =-1, \ \ \alpha =-2, \ \ p= 2-6-7=-11, \ p=2*4-12-7=-11

I assume that we are after two different roots so the solution for p is p=-11

b) \alpha +2 =-2+2=0 \ and \ \beta+2=-1+2=1

So a quadratic equation with the expected roots  is

x(x-1)=x^2-x

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

3 0
3 years ago
Heme d hey the girl who helped me please hl
mestny [16]

Answer:

Parallel segments → 1

perpendicular segments → 2

congruent segments → 3

---------

hope it helps...

have a great day!!

5 0
3 years ago
Read 2 more answers
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