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6 months ago

The length of a rectangle is 4 in longer than its width.

Mathematics

1 answer:

bija089 [108]6 months ago

5 0

The area of the rectangle that has a perimeter of 44 in is calculated as: 117 in².

<h3>What is the Area and Perimeter of a Rectangle?</h3>

Area = length × width

Perimeter = 2(length + width).

Given the following:

Width = w in.
Length = w + 4 in
Perimeter of the rectangle = 44 in.

Therefore:

2(w + 4 + w) = 44

Solve for w

2(2w + 4) = 44

4w + 8 = 44

4w = 44 - 8

4w = 36

w = 9

Width of the rectangle = w = 9 in.

Length = w + 4 = 9 + 4 = 13 in.

Area of the rectangle = 13 × 9 = 117 in².

Learn more about the area of rectangle on:

brainly.com/question/2607596

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What is 30=8+4(z-2) step by step?

Stolb23 [73]

30=8+4(z-2)

Distribute 4 through the parentheses

30=8+4z-8

Eliminate the opposites

30=4z

Swap the sides of the equation

4z=30

Divide both sides of the equation by 4

4z÷4=30÷4

Any expression divided by itself equals 1

z=30÷4

or write the division as a fraction

z=30/4

copy the numerator and denominator of the fraction

30=2x3x5

4=2x2

Write the prime factorization of 30

Write the prime factorization of 4

30=2 x3x5

4=2x2

Line up the common factors in both lists

Copy the common factors

Since there is only one common factor, the common factor 2 is also the greatest common factor

30÷2/4÷2

Divide 30 and 4 by the greatest common factor 2

15/4÷2

Divide the numbers in the numerator

15/2

Divide the numbers in the denominator

15/2

The simplified expression is 15/2

That's it. hope it wasn't too hard to understand?

7 0

2 years ago

Evaluate the integral, show all steps please!

Aloiza [94]

Answer:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

<u>Integration by substitution</u>

<u />

<u /> $\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

<u>Substitute</u> everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$