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3 years ago

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The International Space Station orbits Earth at an altitude of about 240 miles. In the diagram, the Space Station is at point E.

The radius of Earth is approximately 3,960 miles. To the nearest ten miles, what is EH, the distance from the space station to the horizon?

1 answer:

Troyanec [42]3 years ago

4 0

Answer:

are there options ?

Step-by-step explanation:

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5/6, 1/4, and 2/3 yard. Needs 1 yd of cable

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1. So first of all we have to divide the 3 fractions into decimals to get a decimal to compare. So 5/6 is the same as 5 divided by 6 which is .83 bar and so on...

5/6= .83 bar
1/4= .25
2/3= .66 bar

So 5/6 and 2/3 are closer to one.

2. The two shortest pieces are 2/3 and 1/4 so you se 1/4 + 2/3. Let’s get a common denominator for these fractions. The common denominator is 12. So multiply 1/4 • 3 to get 3/12 and multiply 2/3 by • 4 to get 8/12. Add them together and you get 11/12. So he would need 1/12 more cable or 0.083 bar.

3. Now we have to find a common denominator for all of them. The common denominator is 12 again. Multiply 5/6•2 and then you get 10/12, then add 10/12 +2/12(from 1/4) and then you get leftover with 9/12 or 3/4 more wire.

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If a man drives 600km from Johannesburg to Durban. He starts his journey with a full tank of petrol. If the man drives at an ave

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I think it's 5

Step-by-step explanation:

im not sure

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3 years ago

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Given: x + y = 15. Which of the following are ordered pairs of this equation?

Dmitry_Shevchenko [17]

(0,5)(1,4)(2,3) I got those

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3 years ago

Read 2 more answers

Help pls, will mark brainliest

BaLLatris [955]

Here , we are provided with a table which shows 5 consecutive terms of an arithmetic sequence . But before solving further , let's recall that ;

The n'th term of a Arithmetic Sequence let's say it be ${\sf T_n}$ is given by ;

${\boxed{\bf T_{n}=T_{1}+(n-1)d}}$

Where , <u>d</u> is the common difference

Now , here we are given with ;

${\quad \qquad \sf \blacktriangleright \blacktriangleright \blacktriangleright T_{1}=8 \: and \: T_{5}=-4}$

We have to find the 2nd , 3rd and 4th term respectively ,

Now , by using the above formula , 5th term can be written as ;

${: \implies \quad \sf T_{1}+(5-1)d=T_{5}}$

Putting the values and transposing 1st term to RHS , we have ;

${: \implies \quad \sf 4d = -4-8}$

${: \implies \quad \sf d=-\dfrac{12}{4}}$

${: \implies \quad \sf d=-3}$

Now , as we got the common difference , so we can find out the missing terms now ;

${: \implies \quad \sf T_{2}= T_{1}+(2-1)d}$

${: \implies \quad \sf T_{2}= 8 +d}$

${: \implies \quad \sf T_{2}= 8-3}$

${: \implies \quad \bf \therefore \: T_{2}= 5}$

Now

${: \implies \quad \sf T_{3}= T_{1}+(3-1)d}$

${: \implies \quad \sf T_{3}= 8 +2d}$

${: \implies \quad \sf T_{3}= 8-6}$

${: \implies \quad \bf \therefore \: T_{3}= 2}$

Also ,

${: \implies \quad \sf T_{4}= T_{1}+(4-1)d}$

${: \implies \quad \sf T_{4}= 8 +3d}$

${: \implies \quad \sf T_{4}= 8-9}$

${: \implies \quad \bf \therefore \: T_{4}= -1}$

Now , The given table can be written as ;

${\begin{array}{|c|c|c|c|c|c|}\cline{1-6} \bf n & \sf 1 & 2 & 3 & 4 & 5 \\ \cline{1-6} \bf T_{n} & \sf 8 & 5 & 2 & -1 & -4 \end{array}}$

Note :- Kindly view the answer from web , if you're not able to see the full answer from here ;

brainly.com/question/26750175

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3 years ago