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kodGreya [7K]
3 years ago
5

Please help — Which ratio represents cos(c)?

Mathematics
1 answer:
lions [1.4K]3 years ago
5 0
48/50

option A is correct
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What is the square root of 196?
mario62 [17]

the root of 196 is 14.......

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3 years ago
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Unit 2 (12 points)
klio [65]
1+3=4

A counterexample is an a example that proves the statement false.

1+3 are not even numbers but they equal an even one, so it just proved the statement wrong.
3 0
3 years ago
Solve for c.<br><br> 3 − 7c − 20c = 7(–7c − 19) + 14c<br><br> c =
OleMash [197]

Answer:

\boxed{c = -17}

Step-by-step explanation:

=  > 3 - 7c - 20c = 7( - 7c - 19) + 14c \\  \\  =  > 3 - 27c = ( - 7c \times 7) - (7 \times 19) + 14c \\  \\  =  > 3 - 27c =  - 49c - 133 + 14c \\  \\  =  > 3 - 27c =  - 35c - 133 \\  \\  =  > 3 - 27c + 35c =  - 133 \\  \\  =  > 3  + 8c =  - 133 \\  \\  =  > 8c =  - 133 - 3 \\  \\  =  > 8c =  - 136 \\  \\  =  > c =  -  \frac{136}{8}  \\  \\  =  > c =  - 17

7 0
3 years ago
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Evaluate: 2x +3y = 4 -7x + 3y = -1
jolli1 [7]
We solving for x or y
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3 years ago
Find the following limit or state that it does not exist. ModifyingBelow lim With x right arrow minus 2 StartFraction 3 (2 x min
leva [86]

Answer:

-60

Step-by-step explanation:

The objective is to state whether or not the following limit exists

                                \lim_{x \to -2}  \frac{3(2x-1)^2 - 75}{x+2}.

First, we simplify the expression in the numerator of the fraction.

3(2x-1)^2 -75 = 3(4x^2 - 4x +1) -75 = 12x^2 - 12x + 3 - 75 = 12x^2 - 12x -72

Now, we obtain

                         12(x^2-x-6) = 12(x+2)(x-3)

and the fraction is transformed into

                       \frac{3(2x-1)^2 - 75}{x+2} =  \frac{12(x+2)(x-3)}{x+2} = 12 (x-3)

Therefore, the following limit is

       \lim_{x \to -2}  \frac{3(2x-1)^2 - 75}{x+2} = \lim_{x \to -2}  12(x-3) = 12 \lim_{x \to -2} (x-3)

You can plug in -2 in the equation, hence

                        12 \lim_{x \to -2} (x-3) = 12 (-2-3) = -60

6 0
3 years ago
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