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VLD [36.1K]
2 years ago
12

HELP PLS!! C= -3 5c + 4c + 1c=

Mathematics
2 answers:
Airida [17]2 years ago
8 0

Answer:

Solution

If C=-3

5×-3+4×-3+1×-3

= -30

I think it will help you.

7nadin3 [17]2 years ago
7 0

Answer:

-30

Step-by-step explanation:

We are going to substitute c as -3 in all the places it is shown:

(x is just a variable I used to represent the answer)

5c + 4c + 1c = x

5(-3) + 4(-3) + 1(-3) = x

(-15) + (-12) + (-3) = x

-30 = x

The answer is -30

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I need help with finding the answer to a) and b). Thank you!
shtirl [24]

Answer:

\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }

\displaystyle \cos\Big(\frac{x}{2}\Big)=-\frac{3 \sqrt{58}}{58}

\displaystyle \tan\Big(\frac{x}{2}\Big)=-\frac{7}{3}

Step-by-step explanation:

We are given that:

\displaystyle \sin(x)=-\frac{21}{29}

Where x is in QIII.

First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:

a=\sqrt{29^2-21^2}=20

So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.

Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.

And if x is in QIII, this means that:

180

So:

\displaystyle 90 < \frac{x}{2} < 135

Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.

1)

Recall that:

\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}

Since x/2 is in QII, this will be positive.

Using the above information, cos(x) is -20/29. Therefore:

\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 +  20/29}{2}

Simplify:

\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}

2)

Likewise:

\displaystyle  \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }

Since x/2 is in QII, this will be negative.

Using the above information, cos(x) is -20/29. Therefore:

\displaystyle  \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }

Simplify:

\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}

3)

Finally:

\displaystyle \tan\Big(\frac{x}{2}\Big) = \frac{\sin(x/2)}{\cos(x/2)}

Therefore:

\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}

5 0
3 years ago
If f(x)=-3x+2, find<br> -<br> f(-1)
Y_Kistochka [10]

Answer:

The answer is 5

Step-by-step explanation:

f(-1) = -3(-1) + 2

f(-1) = 3 + 2

f(-1) = 5

4 0
3 years ago
) Chứng minh rằng giá trị của biểu thức ((x-2√x)÷(√x-2) ) - ((x-4)÷√x+2)) không phụ thuộc vào giá trị của biến.
Mashutka [201]

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8 0
3 years ago
Find the smallest positive integer k such that 360k is a cube number
Gnoma [55]
Hello,

As 360=2^3*3^2*5,

360k=n^3 <==>k=3*5^2*a^3=75*a^3

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4 years ago
How do you find a missing number in a multiplication equation?
Slav-nsk [51]
If it gives you your total, and another number, divide the 2

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25 • x = 75

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8 0
4 years ago
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