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3 years ago

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The side lengths of a triangle are 5,10, and 13. Is this a right Triangle?

1 answer:

ollegr [7]3 years ago

3 0

No because 5^2 +10^2 is not equavalent to 13^2

by using the pythagoras' theorem

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How do I find the sides of a right triangle using sine, cos, or tan, with only the three angles and the hypotenuse

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What is the slope of a line that is parallel to y = 3x + 5?

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2 years ago

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The results of a test were the following scores: 67, 82, 77, 92, 96, 58, 72, 78, 79, and 82. What is the range for the test scor

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3 years ago

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How do you determine the area under a curve in calculus using integrals or the limit definition of integrals?

RSB [31]

Answer:

Please check the explanation.

Step-by-step explanation:

Let us consider

$y = f(x)$

To find the area under the curve $y = f(x)$ between $x = a$ and $x = b$ , all we need is to integrate $y = f(x)$ between the limits of $a$ and $b$ .

For example, the area between the curve y = x² - 4 and the x-axis on an interval [2, -2] can be calculated as:

$A=\int _a^b|f\left(x\right)|dx$

= $\int _{-2}^2\left|x^2-4\right|dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\int _{-2}^2x^2dx-\int _{-2}^24dx$

solving

$\int _{-2}^2x^2dx$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

$=\left[\frac{x^{2+1}}{2+1}\right]^2_{-2}$

$=\left[\frac{x^3}{3}\right]^2_{-2}$

computing the boundaries

$=\frac{16}{3}$

Thus,

$\int _{-2}^2x^2dx=\frac{16}{3}$

similarly solving

$\int _{-2}^24dx$

$\mathrm{Integral\:of\:a\:constant}:\quad \int adx=ax$

$=\left[4x\right]^2_{-2}$

computing the boundaries

$=16$

Thus,

$\int _{-2}^24dx=16$

Therefore, the expression becomes

$A=\int _a^b|f\left(x\right)|dx=\int _{-2}^2x^2dx-\int _{-2}^24dx$

$=\frac{16}{3}-16$

$=-\frac{32}{3}$

$=-10.67$ square units

Thus, the area under a curve is -10.67 square units

The area is negative because it is below the x-axis. Please check the attached figure.

6 0

3 years ago