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3 years ago

Does anyone know the answer to this?

Mathematics

2 answers:

aalyn [17]3 years ago

8 0

First you need to distribute : -4x + 8x - 3

Combine like terms : 4x -3

STALIN [3.7K]3 years ago

5 0

8. Im assuming you want to:-

Simplify -2x(2 - 4x ) - 3

Distributing the -2x over the parentheses:-

= -4x + 8x^2 - 3

= 8x^2 - 4x - 3

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What is the measure of c in the parallelogram shown

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3 years ago

25% of 60 75% of 30 50% of 45.7<br> 50% of 60 100% of 22.5<br> 75% of 60 10% of 22.5

alexdok [17]

Answer:

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Step-by-step explanation:

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2 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