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dalvyx [7]
4 years ago
6

Using the side splitter theorem, Daniel wrote a proportion for the segments formed by line segment DE. What is EC ?

Mathematics
2 answers:
Sidana [21]4 years ago
7 0
Pls. see attachment.

Luden [163]4 years ago
7 0

Answer:

2.4 units

Step-by-step explanation:

took the quiz

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Let f(x)=6x^2-ax^3 what is the value of a if f has a point of inflection at X=6
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The results of a color spinner experiment are shown in the table. Consider the experimental probability of the spinner landing o
inessss [21]

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10

Step-by-step explanation:

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What number must multiply each side of the equation 25x=10 to produce the equivalent equation x = 25?
insens350 [35]

Answer:

Multiply both sides by \frac{5}{2}

Step-by-step explanation:

Given

\frac{2}{5}x = 10

Required

Get an equivalent of x = 25

To do this, we simply multiply through by \frac{5}{2}

\frac{5}{2} * \frac{2}{5}x = 10 * \frac{5}{2}

\frac{5*2}{2*5} x = 10 * \frac{5}{2}

\frac{10}{10} x = 10 * \frac{5}{2}

\frac{1}{1} x = 10 * \frac{5}{2}

x = 10 * \frac{5}{2}

x =\frac{10*5}{2}

x =\frac{50}{2}

x = 25

5 0
3 years ago
What are the answers for 2, 3 , and 4? That’s it
PIT_PIT [208]

Answer:

1. Formula: b * h ÷ 2

12 * 3 = 36 ÷ 2 = 18 in^2

2. Formula: b * h ÷ 2

5 * 8 = 40 ÷ 2 = 20 ft^2

I could only help on two. Hope it helps though! =)

5 0
3 years ago
The harmonic motion of a particle is given by f(t) = 2 cos(3t) + 3 sin(2t), 0 ≤ t ≤ 8. (a) When is the position function decreas
iren [92.7K]

For the last part, you have to find where f'(t) attains its maximum over 0\le t\le8. We have

f'(t)=-6\sin3t+6\cos2t

so that

f''(t)=-18\cos3t-12\sin2t

with critical points at t such that

-18\cos3t-12\sin2t=0

3\cos3t+2\sin2t=0

3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0

\cos t(3\cos^2t-9\sin^2t+4\sin t)=0

\cos t(12\sin^2t-4\sin t-3)=0

So either

\cos t=0\implies t=\dfrac{(2n+1)\pi}2

or

12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi

where n is any integer. We get 8 solutions over the given interval with n=0,1,2 from the first set of solutions, n=0,1 from the set of solutions where \sin t=\dfrac{1+\sqrt{10}}6, and n=1 from the set of solutions where \sin t=\dfrac{1-\sqrt{10}}6. They are approximately

\dfrac\pi2\approx2

\dfrac{3\pi}2\approx5

\dfrac{5\pi}2\approx8

\sin^{-1}\dfrac{1+\sqrt{10}}6\approx1

2\pi+\sin^{-1}\dfrac{1+\sqrt{10}}6\approx7

2\pi+\sin^{-1}\dfrac{1-\sqrt{10}}6\approx6

4 0
3 years ago
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