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2 years ago

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A beacon is flashing on top of a 50 foot tower. A 6 foot tall man walks constantly away from the tower at 5 feet/sec. At the ins

tant the man is 30 feet away from the tower, what rate is the tip of his shadow moving away from the tower

1 answer:

Viefleur [7K]2 years ago

7 0

Answer: $\frac{253}{44}$

Step-by-step explanation:

ignore the "at the instant the man is 30 feet away" part, set it as X and the man's shadow as Y.

Similar triangles so we can do $\frac{50}{x+y} = \frac{6}{y}$ .

Solve for it we get 44y = 6x

Differentiate relative to time t, we get 44y' = 6x'.

change in x (x') is equal to 5. And we get the answer y' = $\frac{33}{44}$ .

the $\frac{33}{44}$ ft/sec is the rate of which the length of the shadow is changing. add 5 to it for the rate of the tip of his shadow moving away from the tower.

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What is the prime factorization of 108

Anika [276]

☆What is the prime factorization of 108?

To find the prime factorization, first divide 108 by 2.

$108 \div 2 = 54$

You have 2 numbers: 54 and 2. 2 is a prime number and 54 isn't. Divide 54 by 2 until every factor of 54 is prime.

★ Prime number collection: 2

$54 \div 2 = 27$

Add 2 to the "prime number collection". Divide 27 by factors until every factor you find is prime.

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Add 3 to the "prime number collection". Divide 9 by a factor of it to find more prime numbers.

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4 years ago

The acceleration of an object (in m/s^2) is given by the function a(t) = 9 sin(t). The initial velocity of the object is v(0) =

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a) Acceleration is the derivative of velocity. By the fundamental theorem of calculus,

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so that

$v(t)=\left(-11\frac{\rm m}{\rm s}\right)+\int_0^t9\sin u\,\mathrm du$

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b) We get the displacement by integrating the velocity function like above. Assume the object starts at the origin, so that its initial position is $s(0)=0\,\mathrm m$ . Then its displacement over the time interval [0, 3] is

$s(0)+\displaystyle\int_0^3v(t)\,\mathrm dt=-\int_0^3(2+9\cos t)\,\mathrm dt=\boxed{-6-9\sin3}$

c) The total distance traveled is the integral of the absolute value of the velocity function:

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$v(t)$ for $0\le t$ and $v(t)\ge0$ for $\cos^{-1}\left(-\frac29\right)\le t\le3$ , so we split the integral into two as

$\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}-v(t)\,\mathrm dt+\int_{\cos^{-1}\left(-\frac29\right)}^3v(t)\,\mathrm dt$

$=\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}(2+9\cos t)\,\mathrm dt-\int_{\cos^{-1}\left(-\frac29\right)}^3(2+9\cos t)\,\mathrm dt$

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3 years ago

David bought a computer that was 20%

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Answer:

<em>The total price David paid for the computer is $933.12</em>

<em />

Step-by-step explanation:

The cost of the computer will be $1,080 with 20% off and 8% sales tax.

20% of 1080 is 1080(0.20)=216 this means 216 is 20% of 1080, so with the discount the price will be 1080-216=$864

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2 years ago

Kelli is running a mile in gym class. She runs at a pace of 6 minutes per mile. Represent this situation with an equation. Defin

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answer : t= 6d

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We change the speed to miles per minute

6 minutes per mile = 1/6 miles per minute

d is the distance that Kelli has run and t is the number of minutes passed.

Speed is 1/6 mile per minute

We know the formula distance = speed * time

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So the equation becomes

$d= \frac{1}{6}t$

Multiply by 6 on both sides

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4*4 = 16 so,

$-\frac{3}{16}$ is the answer

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