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scoundrel [369]
3 years ago
7

I need help:/ I’m in college

Mathematics
1 answer:
lesantik [10]3 years ago
3 0

Step-by-step explanation:

Amount of acid = 14.9% of 331 mL solution

= 0.149×(331 mL)

= 49.3 mL acid

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For any integers x and y, min(x, y) and max(x, y) denote the minimum and the maximum of x and y, respectively. For example, min(
Annette [7]

Answer:

min(10,w)=\left \{ {{w} \quad\text{for } w

Step-by-step explanation:

No value of w is given, so we can only tell you the meaning of <em>min(10, w)</em>:

When w < 10, min(10, w) is w.

When w ≥ 10, 10 is the smaller of the two values, so min(10, w) = 10.

3 0
3 years ago
First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x
e-lub [12.9K]

Answer:

(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

\int x ln(5+x)dx

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

\int (U-5) ln(U)dU

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

\int (pq')=pq-\int qp'

so we must define p, q, p' and q':

p=ln U

p'=\frac{1}{U}dU

q=\frac{U^{2}}{2}-5U

q'=U-5

and now we plug these into the formula:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU

Which simplifies to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU

Which solves to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C

so we can substitute U back, so we get:

\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C

and now we can simplify:

\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

notice how all the constants were combined into one big constant C.

7 0
3 years ago
Wayne is putting tools into a toolbox, one at a time. He wants to include a pair of pliers, a
Nikolay [14]

Answer:

5 different orders. hope this help

6 0
3 years ago
How much heat is released when 12. 0 grams of helium gas condense at 2. 17 K? The latent heat of vaporization of helium is 21 J/
aleksandr82 [10.1K]

The amount of heat released when 12.0g of helium gas condense at 2.17 K is; -250 J

The latent heat of vapourization of a substance is the amount of heat required to effect a change of state of the substance from liquid to gaseous state.

However, since we are required to determine heat released when the helium gas condenses.

The heat of condensation per gram is; -21 J/g.

Therefore, for 12grams, the heat of condensation released is; 12 × -21 = -252 J.

Approximately, -250J.

Read more on latent heat:

brainly.com/question/19863536

7 0
2 years ago
Sasha invests $200 at 5% after 6 months
Umnica [9.8K]
5% of 200 is 10, after six months, so total of $60? not sure though.

8 0
3 years ago
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