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Kay [80]
3 years ago
15

Please help solve this, I solved it on my own but I think my calculations are wrong. here is the problem,

Mathematics
2 answers:
denis-greek [22]3 years ago
5 0

Hi there! Hopefully this helps!

----------------------------------------------------------------------------------------------------------

<h2>Answer: 12</h2><h2>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</h2>

First we need to <em><u>substitute the value of the variable into the expression</u></em> and simplify.

b-(b+a×4c÷4) = 1 - (1 + -6 × 4(2) ÷ 4).

Now we need to solve 1 - (1 + -6 × 4(2) ÷ 4).

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

1 - (1 + -6 × 4(2) ÷ 4)

|

|  First, we cancel out 4 and 4.

\/

1 - (1 - 6 × 2)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Next, we multiply -6 and 2 to get -12.

1 - (1 - 12)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Then we subtract 12 from 1 to get -11.

1 - (-11)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The opposite of -11 is 11.

1 + 11.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

<h2 /><h2>Add 1 and 11 to get, you guessed it, 12!</h2>
Lostsunrise [7]3 years ago
5 0

Answer:

Answer: 12

Hope this helps

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 Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III
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The values of \sin 2x, \cos 2x, \tan 2x.

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It is given that x lies in the III quadrant. It means only tan and cot are positive and others  are negative.

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\sin^2 x+\cos^2 x=1

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Therefore, the required values are \sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}.

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3 years ago
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