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musickatia [10]
3 years ago
12

How would you solve for x X/8=2

Mathematics
1 answer:
Yanka [14]3 years ago
3 0
You would multiply both sides by 8 in order to get the variable, x, by itself. Then the equation would be X=16.
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Rewrite 9cos 4x in terms of cos x.
rosijanka [135]
\bf \qquad \textit{Quad identities}\\\\
sin(4\theta )=
\begin{cases}
8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\
4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )
\end{cases}
\\\\\\
cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\
-------------------------------\\\\
9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1]
\\\\\\
72cos^4(x)-72cos^2(x)+9


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

as far as the previous one on the 2tan(3x)

\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\

\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[  \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]
\\\\\\
2\left[  \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]
\\\\\\

\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]
\\\\\\
2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}
4 0
3 years ago
Select all of the following that are quadratic equations.
kifflom [539]
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5x-7=0 is NOT quadratic.  The highest exponent of x is 1, not 2.
x²+3x-5=0 is quadratic; the highest exponent of x is 2.
x-5=9x+7 is NOT quadratic.  The highest exponent of x is 1, not 2.
x²-x=3x+7 is quadratic; the highest exponent of x is 2.
5 0
4 years ago
If w = -2 and v = -8, which of the following expressions shows the values correctly substituted in for the variables in the expr
Ahat [919]

Answer:

<em>(-2)2 - (-8) + 1</em>

Step-by-step explanation:

When you substitute values into an algebraic equation, you need to input the values exactly how they are given. "w" must go where there are "w"s, and "v" must go where the "v"s are, otherwise you will get an incorrect answer. All values must also carry over their signs--they are one unit--and must be inserted as such, which is why - (-8) is correct, but not - (8) or - 8. (This is because those two negatives will cancel each other out and become +8 when solving). The parenthesis around the values are important because they protect the original values of the variables, which is why (-2)2 is correct. In the case of (-2)2, it also signifies that they are "attached" by multiplication, and when solving would become -4. Without signifying that the variables are separate from the rest of the equation via the parenthesis, it becomes very easy to solve it incorrectly.

5 0
3 years ago
Read 2 more answers
if a motercycle is moving at a constant speed down the highway of 40 km/hr, how long would it the motorcycle to travel 10 km
ahrayia [7]

Answer:

15 minutes

Step-by-step explanation:

First, the motorcycle goes at a speed of 40 km/hr.

The question asks for how long it would take to travel 10 km.

Well, there are 60 minutes in an hour, since we will be translating into minutes.

Also, 10 km is 1/4 of 40 km, so it would make sense that the time length would be 1/4 of an hour as well.

1/4 of 60 minutes is 15 minutes.  So it takes 15 minutes for the motorcycle to travel 10 km.

Now, if all this wordy stuff is too much to comprehend, you can also solve using proportional relationships.

\frac{40km}{60min}=\frac{10km}{xmin}

Now cross multiply:

40km*xmin=10km*60min\\40x=600

Divide both sides by 40:

\frac{40x}{40}=\frac{600}{40}\\x=15

Again, this shows that it wouls take 15 minutes for the motorcycle to travel 10 km.

6 0
3 years ago
2cot^2(x)-7csc(x)+8=0
Crank
Hello here is a solution : 

6 0
4 years ago
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