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3 years ago

15

a builder needs 1,044 nails to finish a project. if the nails come in packages of 58, how many packages should the builder purch

ase

1 answer:

Natali [406]3 years ago

4 0

9514 1404 393

Answer:

18

Step-by-step explanation:

(1044 nails)/(58 nails/pkg) = 18 pkg

The builder should purchase 18 packages of nails.

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2/3 divide by 2/5<br><br> help please

suter [353]

Answer: 1 /3

Explantion:

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3 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})$\end{minipage}}$

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

$\textsf{Use the trigonometric identity}: \quad \tan \theta=\dfrac{\sin \theta}{\cos \theta}$

$\implies \dfrac{\sin \theta}{9 \cos \theta} +\text{C}$

$\textsf{Substitute back in } \sin \theta=\dfrac{x}{3}:$

$\implies \dfrac{x}{9(3 \cos \theta)} +\text{C}$

$\textsf{Substitute back in }3 \cos \theta=\sqrt{9-x^2}:$

$\implies \dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Learn more about integration by substitution here:

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2 years ago

What is the measure of KPN?

Debora [2.8K]

Answer:

angle KPN=95 degree

Step-by-step explanation:

angle KPN = angle JPO (because they are vertically opposite angles)

Now,

angle JPO+angle LOP=180 degree(being co interior angles)

angle JPO + 85 =180

angle JPO =180-85

angle JPO =95

since angle JPO is equal to KPN ,angle KPN is 95 degree

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3 years ago

What is the volume of the cylinder below? A cylinder with a height of 12 millimeters and diameter of 18 millimeters. 108 pi mm3

Artemon [7]

Answer:

V =972 pi mm ^3

Step-by-step explanation:

The diameter is 18 so the radius is 1/2 of the diameter

18/2 = 9

r=9

The volume of a cylinder is

V = pi r^2 h

V = pi (9)^2 12

V =972 pi mm ^3

8 0

3 years ago

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Carolyn sells homemade candles. Write an expression for her total revenue if she sells x candles for $5 each

jok3333 [9.3K]

Answer:

I think it would be X x 5 = Y

Step-by-step explanation:

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3 years ago