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3 years ago

The length of a rectangle is 4 inches more than the width. the perimeter is 32 inches. find the length and width.

1 answer:

7 0

The length of the rectangle is 10 inches, and the width is 6 inches.

10 + 10 + 6 + 6 = 32

And that's your answer! ^ I hope this helped! c:

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Factorise (x + 2)^2 + p^2 + 2p (x + 2)

Sveta_85 [38]

Answer:

( x +2 +p) ( x+2+p)

Step-by-step explanation:

let x+2 be x

x^2 + p^2 + 2px

x^2 + 2 × p × x + p^2

(x+p)^2

(x+p)(x+p)

replacing the value of x

( x +2 +p) ( x+2+p)

8 0

3 years ago

A gallon of paint will cover 144 square feet of floor space. What are the dimensions of the largest square floor that the gallon

Mkey [24]

You want the largest SQUARE floor then both of the sides have to be the same. Since one gallon can cover up 144 square feet, to work out the dimensions you have to find the square root of 144 which is 12. So the largest square floor it can cover has dimensions 12x12 feet.
Hope this helped :)

8 0

3 years ago

Two fractions between 2 and 2 1/2

Roman55 [17]

Answer:

2 2/5 and 2 2/100

Step-by-step explanation:

The answers range in many ways! Whatever floats your boat to pick!

Hope this helped! :)

7 0

3 years ago

Prove the following integration formula:

7nadin3 [17]

Answer:

See Explanation.

General Formulas and Concepts:

Pre-Algebra

Distributive Property
Equality Properties

Algebra I

Combining Like Terms
Factoring

Calculus

Derivative 1: $\frac{d}{dx} [e^u]=u'e^u$
Integration Constant C
Integral 1: $\int {e^x} \, dx = e^x + C$
Integral 2: $\int {sin(x)} \, dx = -cos(x) + C$
Integral 3: $\int {cos(x)} \, dx = sin(x) + C$
Integral Rule 1: $\int {cf(x)} \, dx = c \int {f(x)} \, dx$
Integration by Parts: $\int {u} \, dv = uv - \int {v} \, du$
[IBP] LIPET: Logs, Inverses, Polynomials, Exponents, Trig

Step-by-step Explanation:

Step 1: Define Integral

$\int {e^{au}sin(bu)} \, du$

Step 2: Identify Variables Pt. 1

Using LIPET, we determine the variables for IBP.

Use Int Rules 2 + 3.

$u = e^{au}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = sin(bu)du\\du = ae^{au}du \ \ \ \ \ \ \ \ \ v = \frac{-cos(bu)}{b}$

Step 3: Integrate Pt. 1

Integrate [IBP]: $\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} - \int ({ae^{au} \cdot \frac{-cos(bu)}{b} }) \, du$
Integrate [Int Rule 1]: $\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{a}{b} \int ({e^{au}cos(bu)}) \, du$

Step 4: Identify Variables Pt. 2

Using LIPET, we determine the variables for the 2nd IBP.

Use Int Rules 2 + 3.

$u = e^{au}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = cos(bu)du\\du = ae^{au}du \ \ \ \ \ \ \ \ \ v = \frac{sin(bu)}{b}$

Step 5: Integrate Pt. 2

Integrate [IBP]: $\int {e^{au}cos(bu)} \, du = \frac{e^{au}sin(bu)}{b} - \int ({ae^{au} \cdot \frac{sin(bu)}{b} }) \, du$
Integrate [Int Rule 1]: $\int {e^{au}cos(bu)} \, du = \frac{e^{au}sin(bu)}{b} - \frac{a}{b} \int ({e^{au} sin(bu)}) \, du$

Step 6: Integrate Pt. 3

Integrate [Alg - Back substitute]: $\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{a}{b} [\frac{e^{au}sin(bu)}{b} - \frac{a}{b} \int ({e^{au} sin(bu)}) \, du]$
[Integral - Alg] Distribute Brackets: $\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2} - \frac{a^2}{b^2} \int ({e^{au} sin(bu)}) \, du$
[Integral - Alg] Isolate Original Terms: $\int {e^{au}sin(bu)} \, du + \frac{a^2}{b^2} \int ({e^{au} sin(bu)}) \, du= \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}$
[Integral - Alg] Rewrite: $(\frac{a^2}{b^2} +1)\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}$
[Integral - Alg] Isolate Original: $\int {e^{au}sin(bu)} \, du = \frac{\frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}}{\frac{a^2}{b^2} +1}$
[Integral - Alg] Rewrite Fraction: $\int {e^{au}sin(bu)} \, du = \frac{\frac{-be^{au}cos(bu)}{b^2} + \frac{ae^{au}sin(bu)}{b^2}}{\frac{a^2}{b^2} +\frac{b^2}{b^2} }$
[Integral - Alg] Combine Like Terms: $\int {e^{au}sin(bu)} \, du = \frac{\frac{ae^{au}sin(bu)-be^{au}cos(bu)}{b^2} }{\frac{a^2+b^2}{b^2} }$
[Integral - Alg] Divide: $\int {e^{au}sin(bu)} \, du = \frac{ae^{au}sin(bu) - be^{au}cos(bu)}{b^2} \cdot \frac{b^2}{a^2 + b^2}$
[Integral - Alg] Multiply: $\int {e^{au}sin(bu)} \, du = \frac{1}{a^2+b^2} [ae^{au}sin(bu) - be^{au}cos(bu)]$
[Integral - Alg] Factor: $\int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)]$
[Integral] Integration Constant: $\int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)] + C$