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goldfiish [28.3K]
3 years ago
13

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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sattari [20]

Answer:

34

Step-by-step explanation:

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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

factors of 19 = 1, 19

factors of 38 = 1, 2, 19 and 38

divide both numerator and denominator by 19

So,

19 / 38 = 1/2

4. 10/18

factors of 10 = 1, 2, 5 and 10

factors of 18 = 1, 2, 3, 6, 9 and 18

So,

10/18

divide both numerator and denominator by 2

= 5/9

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7 0
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:
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