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3 years ago

Kgms^-1 ÷ m× (ms^-1)^2 Can u simplify this

Mathematics

1 answer:

Ymorist [56]3 years ago

7 0

Answer:

(kgm^2) / (s^3)

Step-by-step explanation:

Comment if you want steps.

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A youth group is planning a trip to a theme park. The bus holds up to 40 people. The cost for bus parking is $60.00. Each person

Katena32 [7]

Answer:

C. 0 ≤ X ≤ 40

Step-by-step explanation:

The equation is valid for positive values of x not greater than 40.

_____

Personally, I might put the domain at 1 ≤ x ≤ 40, because there's no point in using the equation (or parking the bus) if zero people are going on the trip.

3 0

3 years ago

What is the answer to (6x-2)-(-x+5)

UNO [17]

Answer:

(6x-2) - (-x+5)

= 6x-2 +x -5

= 7x -7

if you're trying to solve for x, then look below

7x=7

x=1

4 0

3 years ago

Layla has the following data:

ratelena [41]

The answer would be 25 because to find the range you have to subtract the highest number with the lowest number which would be 29-25=4

6 0

3 years ago

Carlos has $500 in his savings account that pays 3% simple interest. What is the amount of money Carlos will have in his bank ac

sleet_krkn [62]

Answer:

$1766

Step-by-step explanation:

hope this helps

4 0

3 years ago

Read 2 more answers

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

3 years ago