This equation has to be written in expanded form -2(-6x

Dina has a mass of 50 kilograms and is waiting at the top of a ski slope that’s 5.0 meters high. What is her potential energy

Bess [88]

Answer:

C

Step-by-step explanation:

Formula for potential energy is PE = m*g*h

where

m is the mass, in kg

g is the acceleration due to gravity, in meters per second squared, and

h is the height, in meters

The problem gives m = 50, g = 9.8, and h = 5. We plug that into the formula and get out answer. So:

$PE=mgh\\PE=(50)(9.8)(5)\\PE=2450$

Correct answer is C.

3 0

4 years ago

Read 2 more answers

What does reciprocal mean in maths?

DaniilM [7]

Answer:

one of a pair of numbers that, when multiplied together, equal 1. such as

6 * 1/6

Step-by-step explanation:

it always equals one

4 0

3 years ago

[(30 + 6) − 32] ÷ 9 ⋅ 2?

Alex777 [14]

The answer is 0.88889, because:
Order of Operations-
1) Add 30 plus 6 to get 36
2) Subtract 32 from 36 to get 4
3) Divide 4 by 9 to get 0.4444444
4) Multiply 0.4444444 by 2 to get 0.888889
Good luck! <3

3 0

4 years ago

<img src="https://tex.z-dn.net/?f=%20%5Ctiny%5Cint_%7Be%7D%5E%7B%7Be%7D%5E%7B2%7D%7D%20%5Cleft%28%20%5Csqrt%7Bx%20%2B%20%5Cfrac%

dlinn [17]

We have the identity

$\left(\sqrt x + \dfrac1{2^n}\right)^2 = x + \dfrac1{2^{n-1}} \sqrt x + \dfrac1{2^{2n}}$

Take the square root of both sides and rearrange terms on the right to get

$\sqrt x + \dfrac1{2^n} = \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt{x} + \dfrac1{2^{n+1}}\right)}$

Decrementing n gives

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \left(\sqrt{x} + \dfrac1{2^{n}}\right)}$

and substituting the previous expression into this, we have

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right) } }$

Continuing in this fashion, after k steps we would have

$\sqrt x + \dfrac1{2^{n-k}} = \sqrt{x + \dfrac1{2^{n-(k+1)}} \sqrt{x + \dfrac1{2^{n-k}} \sqrt{x + \dfrac1{2^{n-(k-1)}} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

After a total of n - 2 steps, we arrive at

$\sqrt x + \dfrac14 = \sqrt{x + \dfrac12 \sqrt{x + \dfrac1{2^2} \sqrt{x + \dfrac1{2^3} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

Then as n goes to infinity, the first nested radical converges to √x + 1/4. Similar reasoning can be used to show the other nested radical converges to √x - 1/4. Then the integral reduces to

$\displaystyle \int_e^{e^2} \left(\sqrt x - \frac14\right) + \left(\sqrt x + \frac14\right) \, dx = 2 \int_e^{e^2} \sqrt x \, dx = \boxed{\frac43 \left(e^3 - e^{\frac32}\right)}$

5 0

3 years ago

$1,800 at 2.1% interest for 2 years What is the interest? What is the new balance?

Contact [7]

Interest is $75.60 new balance is $1,875.60

8 0

3 years ago

This equation has to be written in expanded form -2(-6x - 1)