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aev [14]
3 years ago
11

To successfully escape a planet's gravity, a space ship must achieve escape velocity, which for Earth is 11.2 km/s or 25,054 mil

es per hour. If a ship reaches this speed and maintains it, then neglecting the need to slow down prior to landing, how long would it take the ship to reach the Moon? Use the actual distance to the Moon of 233,815 miles and give your answer in hours.
Mathematics
1 answer:
melisa1 [442]3 years ago
4 0

Answer:

9.33 Hours.

Step-by-step explanation:

The time in hours is given by the distance divided by velocity.

t=D/v

Now we are told that the velocity of the ship will be 25054 miles per hour, and the distance to the Moon is 233815 miles; therefore the time it will take the spacecraft to reach the moon will be:

t=233815miles/(25054miles/hr) = 9.33 hours

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Cuál es primer número natural o entero positivo que conoces​
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Answer:

Los números naturales incluyen solo enteros positivos y comienzan desde 1 hasta infinito

5 0
2 years ago
Which property can be used to solve the equation?<br><br> 5 d = 75
timama [110]

Answer:

Division

Step-by-step explanation:

To isolate d and solve the equation, divide 75 by 5.

4 0
2 years ago
Read 2 more answers
Giant is selling Doritos 2 family bags for 7.00 I’m buying them for the class to celebrate great test scores if I buy 13 bags ho
elena55 [62]
$45.50
2 bags = $7
13 bags=?
7÷2=3.5
3.50=1 bag
$3.50 × 13 bags = $45.50 for 13 bags of Doritos
6 0
3 years ago
sherry read 1/3 of the book in 90 minutes then slept for 3 hours. she continued reading after she woke up and finished the rest
Vadim26 [7]

Answer:

You don’t have question

4 0
3 years ago
2 more questions thanks
sergey [27]
These are two questions and two answers.

1) Problem 17.

(i) Determine whether T is continuous at 6061.

For that  you have to compute the value of T at 6061 and the lateral limits of T when x approaches 6061.

a) T(x) = 0.10x if 0 < x ≤ 6061

T (6061) = 0.10(6061) = 606.1

b) limit of Tx when x → 6061.

By the left the limit is the same value of T(x) calculated above.

By the right the limit is calculated using the definition of the function for the next stage: T(x) = 606.10 + 0.18 (x - 6061)

⇒ Limit of T(x) when x → 6061 from the right = 606.10 + 0.18 (6061 - 6061) = 606.10

Since both limits and the value of the function are the same, T is continuous at 6061.

(ii) Determine whether T is continuous at 32,473.

Same procedure.

a) Value at 32,473

T(32,473) = 606.10 + 0.18 (32,473 - 6061) = 5,360.26

b) Limit of T(x) when x → 32,473 from the right

Limit = 5360.26 + 0.26(x - 32,473) = 5360.26

Again, since the two limits and the value of the function have the same value the function is continuos at the x = 32,473.

(iii) If  T had discontinuities, a tax payer that earns an amount very close to the discontinuity can easily approach its incomes to take andvantage of the part that results in lower tax.

2) Problem 18.

a) Statement Sk

You just need to replace n for k:

Sk = 1 + 4 + 7 + ... (3k - 2) = k(3k - 1) / 2

b) Statement S (k+1)

Replace

S(k+1) = 1 + 4 + 7 + ... (3k - 2) + [ 3 (k + 1) - 2 ] = (k+1) [ 3(k+1) - 1] / 2

Simplification:

1 + 4 + 7 + ... + 3k - 2+ 3k + 3 - 2] = (k + 1) (3k + 3 - 1)/2

                 k(3k - 1)/ 2 + (3k + 1) = (k + 1)(3k+2) / 2

Do the operations on the left side and  you will find it can be simplified to k ( 3k +1) (3 k + 2) / 2.

With that you find that the left side equals the right side which is a proof of the validity of the statement by induction.

4 0
2 years ago
Read 2 more answers
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