Ask question

Ask question

All categories

3 years ago

14

Which are the coordinates of the midpoint of the segment connecting(15,-11) and (-10,10)

1 answer:

saveliy_v [14]3 years ago

7 0

Answer:

Step-by-step explanation:

A(x1,y1) B(x2,y2)

M(x,y) => x= $\frac{x1+x2}{2}$ and y= $\frac{y1+y2}{2}$

-----------------------------------------------------------

A(15,-11) B(-10,10)

M( $\frac{15-10}{2}$ , $\frac{-11+10}{2}$ )=( $-\frac{5}{2}$ , $-\frac{1}{2}$ )

You might be interested in

let's one two and three have the following relationships <1and <2 are adjacent right angles <1 and <3 are vertical a

Fed [463]

Answer:

180 Vertical angles are opposite angles that share only a vertex. Since ∠3 is adjacent to both ∠1 and ∠2, this means that ∠3 shares a side and vertex with both of these angles.

This means that ∠3 and ∠1 form a straight line; this makes them a linear pair, which makes their sum 180°.

8 0

2 years ago

Equivalent expression for 7a+21

Mila [183]

Answer:

Step-by-step explanation:

If you factor the expression 7a+21 you can get:

7 × (a+3)

6 0

3 years ago

The number 7 is what percent of 80

xz_007 [3.2K]

Answer:

5.6

Step-by-step explanation:because if you divide correctly it will equal 5.6

3 0

2 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

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each function with the description

LiRa [457]

Answer:

Step-by-step explanation:

i dont know the answer dud sry

6 0

2 years ago