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Basile [38]
2 years ago
10

PLZ ANSWER THIS thanksssss

Mathematics
1 answer:
just olya [345]2 years ago
5 0

Answer:

It should be C) 11+ 3 5/8

Step-by-step explanation:

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If DE = X+1, EF = x-2, DF =15, and E is<br> between D and F. then DE = ?
mart [117]

Answer:

X =8

Step-by-step explanation:

DF = DE + EF

15 = X+1 + X-2

16 = 2X

X = 8

8 0
2 years ago
What is the value of the expression 3x2 – 2y, if x - -2 and y = -4. 0 20 52 O-8 02​
Dafna11 [192]

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3 years ago
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Lostsunrise [7]
X=11 I say .......... ...
3 0
3 years ago
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According to the data in the figure, find the value of x and y.
drek231 [11]

Answer:

This is a educated guess, but x is 80 and y is 100.

Step-by-step explanation:

My logic here is that the triangles are the same because of all the congruent lines and stuff. But from there, you know that both the right and the left side are 40 degrees. You know the total degree of a triangle is 180 so 180 - 40 - 40 is 100. So the final angle is 100. If you look at x, its on the outside, but on a line. If you can imagine a circle, its half, so its a 180 degrees total. Then its 180 = x + 100. So x is 80. And then by that same logic its y = 100.

6 0
2 years ago
Rewrite 9cos 4x in terms of cos x.
rosijanka [135]
\bf \qquad \textit{Quad identities}\\\\&#10;sin(4\theta )=&#10;\begin{cases}&#10;8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\&#10;4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )&#10;\end{cases}&#10;\\\\\\&#10;cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\&#10;-------------------------------\\\\&#10;9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1]&#10;\\\\\\&#10;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}})}\\\\&#10;-------------------------------\\\\

\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[  \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]&#10;\\\\\\&#10;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]&#10;\\\\\\

\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]&#10;\\\\\\&#10;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
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