MATHEMATICS COLLEGE

Help D: An object is dropped from a small plane. As the object falls, its distance, d, above the ground after t seconds, is given by the formula d = –16t2 + 1,000. Which inequality can be used to find the interval of time taken by the object to reach the height greater than 300 feet above the ground?A.-16t^2+1000<300

B. -16t^2+1000<_ 300

C. -16t^2+1000>_300

D.-16t^2+1000>300

Answers

Answer 1

Answer:

Answer:

D. $-16t^2+1000>300$ is the correct answer.

Step-by-step explanation:

It is given that Distance, d above ground with time 't' is given by the formula:

$d = -16t^2+1000$

The negative sign with $16t^2$ indicates that the distance is decreasing with square of time. i.e. value is getting subtracted from a value 1000.

For example, if t = 0, d = 1000 feet

If t = 2, d = -16 $*$ 4 + 1000 = 936 feet

We can clearly see that when 't' is increasing, the distance 'd' is decreasing.

And at a certain time, the object will be on ground when d = 0 feet.

Inequality for the distance greater than 300 feet i.e.

d > 300 feet

Hence, the inequality will be:

$-16t^2+1000 >300$ is the correct answer.

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Answers

Answer:

d no more than two questions correct

A shoe store recorded the price (in dollars) of 175 pairs of women's shoes and 175 pairs of men's shoes. The data collected was used to create the following table.

Answers

Answer:

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Step-by-step explanation:

i did this and got it correct

Find a solution to the following initial-value problem: dy dx = y(y − 2)e x , y (0) = 1.

Answers

This equation is separable, as

$(\mathrm dy)/(\mathrm dx)=y(y-2)e^x\implies(\mathrm dy)/(y(y-2))=e^x\,\mathrm dx$

Integrate both sides; on the left, expand the fraction as

$\frac1{y(y-2)}=\frac12\left(\frac1{y-2}-\frac1y\right)$

Then

$\displaystyle\int(\mathrm dy)/(y(y-2))=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C$

$\implies\frac12\ln\left|\frac{y-2}y\right|=e^x+C$

Since $y(0)=1$ , we get

$\frac12\ln\left|\frac{1-2}1\right|=e^0+C\implies C=-1$

so that the particular solution is

$\frac12\ln\left|\frac{y-2}y\right|=e^x-1\implies\boxed{y=\frac2{1-e^(2e^x-2)}}$

I’m really struggling, someone please help!

Answers

Hi there! :)

Answer:

$\huge\boxed{C.}$

We can examine each answer choice individually:

A. 569 × 10² = 569 × (10 · 10) = 569 · 100 = 56,900. Therefore, this choice is incorrect.

B. 569 · 10 = 5,690. This choice is incorrect.

C. 10³ · 569 = (10 · 10 · 10) ·569 = 1000 · 569 = 569,000. This choice is correct.

D. 10² · 569 = (10 · 10) · 569 = 56,900. This choice is incorrect.

Therefore, the correct option is C.

Answer:

A. 569 × 10² = 569 × (10 · 10) = 569 · 100 = 56,900.

B. 569 · 10 = 5,690.

C. 10³ · 569 = (10 · 10 · 10) ·569 = 1000 · 569 = 569,000.

D. 10² · 569 = (10 · 10) · 569 = 56,900.

So your answer is C

Enter the range of values for x

Answers

Greetings from Brasil...

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5X - 10 < 25

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The AB side can be neither zero nor negative. So

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