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Olegator [25]
2 years ago
9

If y varies directly as x, and y is 20 when x is 4, what is the constant of variation for this relation?

Mathematics
2 answers:
Mekhanik [1.2K]2 years ago
7 0

Answer:

[DATA EXPUNGED]

Step-by-step explanation:

[REDACTED]

solong [7]2 years ago
7 0

Answer:

Step-by-step explanation:their are many vbarity

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3 3/5+ 4 1/8=<br> PLS PLS PLS PLS ANSWER
Jet001 [13]
I think the answer is 7.725 so sorry if I’m wrong but see if that answer is correct
5 0
2 years ago
I wanna pass this help me please-
Simora [160]

Answer:

C

Step-by-step explanation: hope this helps :)

5 0
2 years ago
Which answer is a solution of the equation?
Nuetrik [128]

2x + y = 5

Put the coordinates of the points to the equation and check it:

A. 5 - not make sense

B. (1, 2) - x = 1, y = 2

2(1) + 2 = 2 + 2 = 4 ≠ 5  NOT

C. (2, -1) - x = 2, y = -1

2(2) + (-1) = 4 - 1 = 2 ≠ 5    NOT

D. (3, -1) - x = 3, y = -1

2(3) + (-1) = 6 - 1 = 5    YES

<h3>Answer: D. (3, -1).</h3>
8 0
3 years ago
Write expression as a sine or cosine of an angle.<br>cos 94° cos 18° + sin 94° sin 18° ​
MArishka [77]
<h3>Answer:  cos(76)</h3>

=========================================================

Explanation:

The original expression is of the pattern cos cos + sin sin. This pattern matches the second identity in the hint. Specifically, we'll say the following:

cos(A - B) = cos(A)cos(B) + sin(A)sin(B)

cos(A)cos(B) + sin(A)sin(B) = cos(A - B)

cos(94)cos(18) + sin(94)sin(18) = cos(94 - 18)

cos(94)cos(18) + sin(94)sin(18) = cos(76)

----------

We can verify this by use of a calculator. Make sure your calculator is in degree mode.

  • cos(94)cos(18) + sin(94)sin(18) = 0.24192
  • cos(76) = 0.24192

Both expressions give the same decimal approximation, so this helps confirm the two expressions are equal. You could also use the idea that if x = y, then x-y = 0. Through this method, you'll subtract the left and right hand sides and you should get (very close to) zero.

6 0
2 years ago
AABC ~ AQRS
Stella [2.4K]
It should be 24 ! Lmk
4 0
3 years ago
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