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3 years ago

From 75 pencil to 225 pencils. Round to the nearest percent.

Mathematics

2 answers:

GalinKa [24]3 years ago

7 0

Answer:

200 %

Step-by-step explanation:

From the question above, the pencils increases from 75 to 225, so what we are looking for is the percentage increase

The formula for finding percentage increase is;

% increase = $\frac{Increase}{Original}$ × 100%

First; we will find the difference between the two numbers i.e 75 and 225

Increase = New number - Original number

= 225 - 75

= 150

Then we divide the increase by the original number and then multiply by 100 %

That is;

% increase = $\frac{150}{75}$ × 100%

= 2 × 100%

= 200%

Therefore; the percentage increase is 200%

MatroZZZ [7]3 years ago

4 0

For finding percents, I use a method called FOO
which is final-original÷original
the original is the first number and the final is the second
After FOO you must multiply by 100 to make it a percent
The equation is as followed:
225-75÷75×100
The answer is 125%

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4 years ago

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Answer:

Area A = 12 = 1/2 × x × (5x-2)

5x^2 - 2x -24 = 0

Solution;

Base x = 2.4

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Step-by-step explanation:

Given;

base of the triangle is x feet;

Base = x

The height of a triangle is 2 less than 5 times it's base;

Height = 5x - 2

the area of the triangle is 12 square feet;

Area A = 12 ft^2

The area of a triangle is;

Area A = 1/2 × base × height

Substituting the base and height;;

Area A = 1/2 × x × (5x-2)

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12 = 1/2 × (5x^2 - 2x)

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24 = (5x^2 - 2x)

5x^2 - 2x -24 = 0

Solving the simultaneous equation;

x = 2.4 or -2

Since the base cannot be negative;

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3 years ago

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Step-by-step explanation:

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3 years ago

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serg [7]

Answer:

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

Step-by-step explanation:

We are given the following integral:

$\int \frac{dx}{\sqrt{9-x^2}}$

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

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$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$

Coming back to x:

We have that:

$x = 3\sin{\theta}$

$\sin{\theta} = \frac{x}{3}$

Applying the arcsine(inverse sine) function to both sides, we get that:

$\theta = \arcsin{(\frac{x}{3})}$

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

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