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svlad2 [7]
3 years ago
11

Which type of stored energy is released when something burns?

Mathematics
1 answer:
krek1111 [17]3 years ago
7 0

Answer:

<h3 /><h3><em><u>CHEMICAL</u></em><em><u> </u></em><em><u>ENERGY</u></em></h3>

Step-by-step explanation:

<em>HOPE </em><em>THIS </em><em>ANSWER</em><em> </em><em>WAS </em><em>HELPFUL</em>

<em>MARK </em><em>AS </em><em>BRAINLIST </em><em>PLEASE</em>

<em>IF </em><em>THIS </em><em>ANSWER</em><em> </em><em>WAS</em><em> </em><em>HELP </em><em>TO</em><em> </em><em>YOU </em>

<em>THANKYOU!</em><em>!</em>

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Peter is reading a 193-page book. he has read three pages more than one-fourth of the number of pages he hasn't yet read. how ma
exis [7]
Let's call <em />the number of pages we <em>have </em>read m. To find the number of pages we have <em>left</em> to read, we'll need to subtract the number of pages we've read from the total number of pages in the book. Our total page count is 193, and the number of pages we've read is the unknown m, so the number of pages we <em />haven't read can be written as 193-m. With that knowledge, let's construct the rest of our equation.

"He has read three pages more than one-fourth of the number of pages he hasn't yet read." The first key phrase here is "he has read." What this means is that the number of pages he's read, which we've named m, is equal to everything in the rest of the sentence. "Thee pages more than" means that we're going to be adding 3 to whatever the next quantity described is, which, in this case, is "one-fourth of the number of pages he hasn't yet read." We already established earlier that "the number of pages he hasn't yet read" is 193-m, so one-fourth of that would be \frac{1}{4} (193-m). Putting everything together, we get the following equation:

m= \frac{1}{4} (193-m)+3

To solve for m, we might first want to eliminate the fraction 1/4 by multiplying both sides of the equation by 4, obtaining

4m=193-m+12

Solving for m, we get:

4m=193+12-m\\5m=205\\m=41

Now that we have m, it's a simple matter to solve for 193-m:

193-41=152

So, he has not yet read 152 pages.
6 0
2 years ago
The following is a model showing the solution to an equation. x - 34 = 13 x - 34 + 34 = 13 + 34 x = 47 What property of equality
yanalaym [24]

Answer:

B.) Addition*Additive property of equality

Step-by-step explanation:

You add 34 to both sides of the equation to find what x was equal to

Hope I helped

8 0
3 years ago
Which unit would you use to measure the weight of an aspirin?<br><br> mg<br> g<br> kg<br> km
Sergeeva-Olga [200]
Milligrams I'm not sure
6 0
3 years ago
Read 2 more answers
B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =
jolli1 [7]

I suppose you mean

g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)

Differentiate one term at a time.

Rewrite the first term as

\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}

Then the product rule says

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'

Then with the power and chain rules,

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}

Simplify this a bit by factoring out \frac12 (36-x^2)^{-3/2} :

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}

For the second term, recall that

\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}

Then by the chain rule,

\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}

So we have

g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}

and we can simplify this by factoring out 18(36-x^2)^{-3/2} to end up with

g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}

5 0
2 years ago
Can somebody help me with this pleaseee
AysviL [449]

Answer:

lesser x = -4.73205, so -4.73

greater x = -1.26795, so -1.27

7 0
2 years ago
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