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2 years ago

Write the product as the product of a whole number and a unit fraction 4 x 2/6

1 answer:

7 0

When you multiply a whole number by a fraction, you multiply the whole number by the numerator (top) of the fraction and you end up with an improper fraction. This question simply wants you to multiply the whole number by the numerator, then multiply that whole number by the remaining fraction.

= 4 * 2/6
multiply 4 by the numerator
=(4*2)/6
multiply 4*2 then separate from the remaining 1/6 fraction
=8 * 1/6

ANSWER: 8 * 1/6

Hope this helps! :)

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At one time during 2001 for every 20 copies of Harry Potter and the Sorcerer's Stone that were sold 13.2 copies of Harry Potter

Airida [17]

In order to solve this, you have to take the number of copies of the Prisoner of Azkaban and divide it by the number of copies of the Sorcerer's Stone:
13.2÷20=0.66
To get this into a percentage, multiply the 0.66 by 100, and get 66%, meaning that The Prisoner of Azkaban was sold only 66% as much as The Sorcerer's Stone.

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3 years ago

Y=-x+5 Y=x+1 Solve the system

Elis [28]

Answer:

$y = - x + 5 \\ y = x + 1 \\ - x + 5 = x + 1 \\ - 2x = - 4 \\ x = 2 \\ y = 2 + 1 = 3 \\ y = 3$

Therefore the answer is:

x=2; y=3

Good luck!

Intelligent Muslim,

From Uzbekistan.

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3 years ago

Can someone help me pls i put rhe question as a picture i would help alot

GuDViN [60]

The answer would be D

8 0

2 years ago

Read 2 more answers

What are the Terms for 8x+6

densk [106]

Answer:Simplifying

8x + 6

Reorder the terms:

6 + 8x

Factor out the Greatest Common Factor (GCF), '2'.

2(3 + 4x)

Final result:

2(3 + 4x)

6 0

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:

Fundamental Theorem of Calculus

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

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:

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

Integration by substitution

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

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

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

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

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

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

Substitute 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})$\end{minipage}}$