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3 years ago

Find the area of the circle with a circumference of 50.24 units help me plz

1 answer:

swat323 years ago

4 0

Circumference of a circle=diameter times pi

50.24=d(3.14)
d=16

radius is half of diameter so radius is 8

Area of a circle=pi(r^2)

(3.14)(8^2)=(3.14)(64)=200.96

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Define the Slope Intercept form of an equation. * hurry

sineoko [7]

Step-by-step explanation:

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1 year ago

Which is the BEST estimate of the average rate of change for the function given in the table, over the interval 2 ≤ x ≤ 4?

ladessa [460]

Answer:

Step-by-step explanation:

Use the slope formula:

$m = \frac{y_2-y_1}{x_2-x_1}$

Picking x = 2 and x = 4 as my two points (since the interval is 2 ≤ x ≤ 4), we get:

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3 years ago

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Solve for e. g/e^2 - r = m - y

vesna_86 [32]

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4 years ago

Write an equation in point-slope form of the line using points D(0,5),E(5,8)

zhenek [66]

Answer:

8 - 5 = m(5 - 0)

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3 years ago

...... trying to find domain, range rel max rel min end behavior inc intervals dec intervals. zeros..... I'm a mom trying to hel

Fiesta28 [93]

Answer:

The values are evaluated below.

Step-by-step explanation:

Given function is

$-x^4-7x^3-12x^2$

We have to find the domain, range, rel max, rel min, end behaviour, increasing or decreasing intervals and zeros of polynomial.

Domain:

The domain of a function is the set of all possible values of x for the given function.

Here domain is set of all real numbers R.

Range:

The range is the resulting y-values we get after substituting all the possible x-values.From the graph we see that

The range is $-\infty$

Relative maxima and minima of a function, are the largest and smallest value of the function on an entire domain of function.

Relative max of $-x^4-7x^3-12x^2$ is 0 at x=0

and $\frac{3}{512}(133\sqrt57-471)$ at $x=\frac{-21}{8}-\frac{\sqrt57}{8}$

Relative min is

$\frac{-3}{512}(471+133\sqrt57)$ at $x=\frac{-21}{8}+\frac{\sqrt57}{8}$

From the graph we see that

End behaviour is As $\\x->\infty, f(x)->-\infty\\x->-\infty, f(x)->-\infty\\$

Increasing intervals and decreasing intervals are

Increasing: $\frac{1}{8}(\sqrt57-21)$

Decreasing: $\frac{1}{8}(-21-\sqrt57)$

Zeroes are :

$-x^4-7x^3-12x^2=0$

⇒ $(-x^2)(x^2+7x+12)=0$

⇒ $(-x^2)(x+3)(x+4)=0$

Hence, zeroes are 0, 0, -3, -4

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3 years ago

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