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natka813 [3]
3 years ago
10

Someone help me pleaseeeeee

Mathematics
2 answers:
Simora [160]3 years ago
4 0

Answer:

option B

Step-by-step explanation:

Top right graph

Lena [83]3 years ago
4 0

Answer:

b

Step-by-step explanation:

pls mark me brainliest

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5. Solve the problem and find the answer
Ghella [55]

Answer:

  x ≈ 263.1

Step-by-step explanation:

The marked sides are the hypotenuse and the side adjacent to the given angle. The relevant trig relation is ...

  Cos = Adjacent/Hypotenuse

  cos(70°) = 90/x . . . . use given values

  x = 90/cos(70°) . . . . multiply by x/cos(70°)

  x ≈ 263.1

_____

<em>Additional comment</em>

The mnemonic SOH CAH TOA is intended to remind you of the primary trig relations:

  Sin = Opposite/Hypotenuse

  Cos = Adjacent/Hypotenuse

  Tan = Opposite/Adjacent

7 0
3 years ago
Which of the sets of ordered pairs represents a function? (5 points) A = {(3, −5), (4, 6), (−3, 9), (2, 7)} B = {(2, 4), (−1, −7
Flauer [41]

Answer: both A and b

5 0
3 years ago
Read 2 more answers
Which substance is detected by a flame test?
Neko [114]
I believe the correct answer is C: gas

8 0
3 years ago
Read 2 more answers
The table shows the portion of students at a middle school that are in each grade. Order the grades from the least to greatest n
juin [17]

Answer:

1. don't see a table so bye 2.

Step-by-step explanation:

8 0
3 years ago
Find all solutions to the equation in the interval [0, 2π).<br><br> cos 4x - cos 2x = 0
Andrew [12]
Solve cos(4x)-cos(2x)=0  &forall; 0<=x<=2pi  ..............(0)

Normal solution:
1. use the double angle formula to decompose, and recall cos^2(x)+sin^2(x)=1
cos(4x)=cos^2(2x)-sin^2(2x)=2cos^2(2x)-1    .................(1)
2. substitute (1) in (0)
2cos^2(2x)-1-cos(2x)=0
3. substitute u=cos(2x)
2u^2-u-1=0
4. Solve for x
factor
(u-1)(u+1/2)=0
=> u=1 or u=-1/2
However, since cos(x) is an even function, so solutions to
{cos(2x)=1, cos(-2x)=1, cos(2x)=-1/2 and cos(-2x)}  ...........(2)
are all solutions.
5. The cosine function is symmetrical about pi, therefore
cos(-2x)=cos(2*pi-2x), 
solution (2) above becomes
{cos(2x)=1, cos(2pi-2x)=1, cos(2x)=-1/2, cos(2pi-2x)=-1/2}
6. Solve each case
cos(2x)=1 => x=0
cos(2pi-2x)=1 => cos(2pi-0)=1 => x=pi
cos(2x)=-1/2 => 2x=2pi/3 or 2x=4pi/3 => x=pi/3 or 2pi/3
cos(2pi-2x)=-1/2 => 2pi-2x=2pi/3 or 2pi-2x=4pi/3 => x=2pi/3 or x=4pi/3
Summing up,
x={0,pi/3, 2pi/3, pi, 4pi/3}
3 0
3 years ago
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