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4 years ago

6

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

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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3 years ago

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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 $\frac{5}{2}$

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

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

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

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

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

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

$x =\frac{50}{2}$

$x = 25$

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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! =)

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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

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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

$\dfrac\pi2\approx2$

$\dfrac{3\pi}2\approx5$

$\dfrac{5\pi}2\approx8$

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

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

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

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3 years ago