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4 years ago

Which is not something that energy can do?

2 answers:

6 0

A. Be completely used up.⇒ NO
B. Cause an object to move.⇒ YES
C. Change an object's temperature.⇒YES
<span>D.Change into other forms of energy</span>⇒<span>.YES</span>

malfutka [58]4 years ago

3 0

A energy can never be completely used up. When something "runs out of energy" it just runs out of use able energy that can be used to do the task that it was doing in the first place so it is now useless to that object and it must move on to be used for a different object.

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Each year the local tax authority issues an appraisal statement, which is a notice of how valuable a property is and at what val

White raven [17]

C. dispute the valuation if he or she disagrees with it.

4 0

3 years ago

Ava, Grace, Jayden, and Zoe together stacked a total of 50 brownies on a tray. Ava stacked 0.14 of the brownies, Grace stacked 4

Karo-lina-s [1.5K]

Total brownies: 50
Ava : 0.14 of brownies
Grace : 4 of the 50 brownies
Jayden : 24% of the brownies
Zoe: remaining

50 - (50 x 0.14) - 4 - (50 x 24%) = 50 - 7 - 4 - 12 = 50 - 23 = 27

Zoe stacked the most brownies. She stacked 27 brownies out of the 50 brownies.

7 0

4 years ago

Four-thirds times the sum of a number and 8 is 24. what is the number?

labwork [276]

So first you take the words and put them into an equation. "a number" that never gets directly told to you will always be a variable, which is easiest to make X. "The sum of a number and 8" will mean that 8+something is the sum so they have to go in parenthesis... giving you
4/3*(x+8)=24
now just solve for X, remembering to use the reverse of PEMDAS, since you're solving for the variable.
remove what is furthest away to the X first.
x+8=24*3/4
x+8=18
x=18-8
x=10
Now, remember to always go through and replace your variable with its value and solve. if it is not equal on both sides, retry the problem because you may have missed a small +, -, or * somewhere!

Hope this helped:)

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Cdisplaystyle%20%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%20%5Cpi%7D%7B2%7D%20%7D%

murzikaleks [220]

Let $x = \arcsin(y)$ , so that

$\sin(x) = y$

$\tan(x)=\dfrac y{\sqrt{1-y^2}}$

$dx = \dfrac{dy}{\sqrt{1-y^2}}$

Then the integral transforms to

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy$

Integrate by parts, taking

$u = \ln(y) \implies du = \dfrac{dy}y$

$dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|$

For 0 < y < 1, we have |1 - y²| = 1 - y², so

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

Recall the Taylor series for ln(1 + y),

$\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n$

Replacing y with -y² gives the Taylor series

$\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}$

and replacing ln(1 - y²) in the integral with its series representation gives

$\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy$

Interchanging the integral and sum (see Fubini's theorem) gives

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy$

Compute the integral:

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}$

and we recognize the famous sum (see Basel's problem),

$\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$

So, the value of our integral is

$\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}$

6 0

3 years ago

FInd the equation of the line

enot [183]

Answer:

y=-3/2x+3

Step-by-step explanation:

The slope is rise over run. It rises 3 and goes back -2 so the slope is -3/2 and it crosses the y-axis at 3 so the y-intercept is 3. The slope equation is y=mx+b.

6 0

3 years ago