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3 years ago

13

Which statements are true about the circle shown? Check all that apply.

1 answer:

otez555 [7]3 years ago

6 0

Step-by-step explanation:

A circle is only composed of the points on the border
The distance around the circle is called circumference

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PLEASE HELP FAST PLEASE AND THANK YOU HELP MY SISTER MEH

Gnoma [55]

The answers are B , D , and F

3 0

3 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago

Help! I have looked at so many lessons on writing exponential functions from a graph and nothing has helped.

Over [174]

Answer:

y = $8(-\frac{1}{2})^x$

Step-by-step explanation:

Let the equation of the exponential function is,

y = a(b)ˣ

Since the graph of this function passes through two points $(4, \frac{1}{2})$ and (3, -1)

For the point $(4, \frac{1}{2})$ ,

$\frac{1}{2}=a(b)^4$ ---------(1)

For second point (3, -1),

-1 = a(b)³ ---------(2)

Divide equation (2) by equation (1),

$\frac{\frac{1}{2}}{-1}$ = $\frac{a(b)^4}{a(b)^3}$

$-\frac{1}{2}=b^{(4-3)}$

$-\frac{1}{2}=b$

From equation (2)

-1 = $a(-\frac{1}{2})^3$

-1 = $-\frac{1}{8}a$

a = 8

Therefore, equation of the exponential function will be,

y = $8(-\frac{1}{2})^x$

4 0

2 years ago

7717 as a percentage

Georgia [21]

The percentage of 7717 is 129.8%

8 0

3 years ago

Enter an equation of the form y= mx for the line that passes through the

hammer [34]

Answer:

y = 13x

Step-by-step explanation:

gradient = 13

x = 0

y = 0

$\frac{Δy}{Δx}$

so $\frac{y - 0}{x - 0}$ = 13

cross multiply and you get

y = 13x

5 0

3 years ago