All categories

3 years ago

The distance from Earth to Mars is 136,000,000 miles. A spaceship travels at 31,000 miles per

2 answers:

7 0

136,000,000/31,000= how many hours for the spaceship to reach mars
136,000,000/31,000/24= how many days for the spaceship to reach mars
136,000,000/31,000/24 is approx. 182.7 days or 183 days (rounded up b/c of "to the nearest day").

Amiraneli [1.4K]3 years ago

7 0

It will take approximately 183 days for them to get to mars. i found this out by first dividing 136,000,000 by 31,000 and i got 4,387.097. i then divided that by 24 (because 24 hour days) i got 182.7956989247312 and just rounded that off to 183 days

You might be interested in

Solve the equation x2 + 7x = 30.

Law Incorporation [45]

Answer:

2*7=14+6=30 points plz

5 0

3 years ago

Please answer this correctly

sineoko [7]

Answer:

Chicken: 36%

Beef: 34%

Black Bean: 30%

Hope this helps!

8 0

3 years ago

A rental car company charges $30 a day plus $0.05 a mile. Which expression could be used to find the cost of renting a car for m

Gnoma [55]

Answer:

Step-by-step explanation:

Since the company earns $0.05 for every mile, we would multiply m by 0.05. We also add it with 30 since it is a base charge. This makes it C

It can't be A or B since that would mean the company earns $30.05 for every mile.

It can't be D since that would mean the company earns $30 for every mile with a base charge of $0.05.

6 0

3 years ago

Put these numbers in order from least to greatest. 0.65 31.56 6.51 6.5 5.16

liberstina [14]

Answer:

0.65

5.16

6.5

6.51

31.56

Step-by-step explanation:

7 0

3 years ago

(x +y)^5 Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$