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madam [21]
4 years ago
7

A bookstore is giving a discount of $9 on paperback books that normally sell for $25. What is the discount rate?

Mathematics
2 answers:
svlad2 [7]4 years ago
7 0
The answer is 36%

~Hope this was of help~
Irina-Kira [14]4 years ago
5 0
The discount would be 74% because the discount is over half off.
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Whitch ordered pair is a solution of y>5xx-2​
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Answer:

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Step-by-step explanation:

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3 years ago
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Find the surface area of the triangular prism below. The bases each have an
Evgen [1.6K]

Answer:

A. 84 cm²

Step-by-step explanation:

Surface area of triangular prism = 2(base area) + (Perimeter)*height of prism

Base area = 6 cm² (given)

Perimeter = 12 cm

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3 years ago
Integration using part formula<br> <img src="https://tex.z-dn.net/?f=%5Cint%5Climits%20%7Bx%5Enlogx%7D%20%5C%2C%20dx" id="TexFor
liq [111]

Answer:

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Step-by-step explanation:

Given integral is I= \int {x^{n}logx \, dx

Take logx=t

x=e^{t}

x^{n}=e^{nt}

\frac{1}{x} dx=dt

dx=xdt

dx=e^{t}dt

I= \int (e^{nt})(t)(e^{t})\, dt

I= \int (e^{(n+1)t})(t)\, dt

Using integration by part,

I= (t)\int [e^{(n+1)t}]\, dt-\int[\frac{d}{dt}{t}\times\int (e^{(n+1)t})]\\\\I= (t) [\frac{1}{n+1}e^{(n+1)t}]-\int[1\times\frac{1}{n+1}e^{(n+1)t}]\,dt\\\\I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]

Writing in terms of x

I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]

I=[\frac{logx}{n+1}e^{(n+1)logx}]-[\frac{1}{(n+1)^{2}}e^{(n+1)logx}]

I=[\frac{logx}{n+1}e^{logx^{(n+1)}}]-[\frac{1}{(n+1)^{2}}e^{logx^{(n+1)}}]

I=[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]

Thus,

Integration of I= \int {x^{n}logx \, dx=[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]

3 0
3 years ago
What’s 1 and what’s 2 ?
katrin2010 [14]

Answer:

For question 1:

The length of hypotenuse expression =√(8^2+9^2)

For question 2:

The expression for length a is

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Step-by-step explanation:

For question 1:

The length of hypotenuse expression =√(8^2+9^2)

For question 2:

Then expression for length a is

a=√(15^2-4^2)

3 0
3 years ago
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