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3 years ago

A trapezoid has an altitude of 12 cm. Its parallel sides measure 10 and 8 cm, respectively. What is its area?

Mathematics

2 answers:

Maurinko [17]3 years ago

8 0

108cm^2 is the answer hope this helps!

stira [4]3 years ago

4 0

Area of trapezoid = 1/2(b1+b2)(h)
<span>1/2(10 cm + 8 cm)(12cm) = </span>

<span>108 cm^2</span>

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3 years ago

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3 years ago

Choose the fraction that has not been reduced to simplest form.

Gennadij [26K]

We are required to choose the fraction that has not been reduced to simplest form

The fraction that has not been reduced to simplest form is 19/38 and 10/18

1. 21 / 34 is in its simplest form

factors of 21 = 1, 7 and 21

factors of 34 = 1, 2, 17 and 34

No common factor of 21 and 34 except 1

2. 15/16

factors of 15 = 1, 3, 5, 15

Factors of 16 = 1, 2, 4, 8 and 16

No common factor of 15 and 16 except 1

3. 19/38

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factors of 38 = 1, 2, 19 and 38

divide both numerator and denominator by 19

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3 years ago

Find the general indefinite integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.

Maru [420]

Answer:

$9\text{ln}|x|+2\sqrt{x}+x+C$

Step-by-step explanation:

We have been an integral $\int \frac{9+\sqrt{x}+x}{x}dx$ . We are asked to find the general solution for the given indefinite integral.

We can rewrite our given integral as:

$\int \frac{9}{x}+\frac{\sqrt{x}}{x}+\frac{x}{x}dx$

$\int \frac{9}{x}+\frac{1}{\sqrt{x}}+1dx$

Now, we will apply the sum rule of integrals as:

$\int \frac{9}{x}dx+\int \frac{1}{\sqrt{x}}dx+\int 1dx$

$9\int \frac{1}{x}dx+\int x^{-\frac{1}{2}}dx+\int 1dx$

Using common integral $\int \frac{1}{x}dx=\text{ln}|x|$ , we will get:

$9\text{ln}|x|+\int x^{-\frac{1}{2}}dx+\int 1dx$

Now, we will use power rule of integrals as:

$9\text{ln}|x|+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\int 1dx$

$9\text{ln}|x|+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\int 1dx$

$9\text{ln}|x|+2x^{\frac{1}{2}}+\int 1dx$

$9\text{ln}|x|+2\sqrt{x}+\int 1dx$

We know that integral of a constant is equal to constant times x, so integral of 1 would be x.

$9\text{ln}|x|+2\sqrt{x}+x+C$

Therefore, our required integral would be $9\text{ln}|x|+2\sqrt{x}+x+C$ .

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2 years ago

I Obviously don't need help but here's a survey answer and you will receive 10 points, Here you go:

sineoko [7]

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3 years ago

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