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3 years ago

What is the midpoint of the mine segment graph below

Mathematics

1 answer:

yanalaym [24]3 years ago

7 0

Answer:

(4, 7/2)

Step-by-step explanation:

Take the mean of the x-coordinates and the mean of the y-coordinates.

x = (-1 + 9)/2 = 8/2 = 4

y = (2 + 5)/2 = 7/2

Answer: (4, 7/2)

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Find the difference. 4 and two-fifths minus 1 and seven-tenths equals what number?

Varvara68 [4.7K]

Answer:

I believe it's B or C.

3 0

3 years ago

Read 2 more answers

równanie zmiany prędkość autokaru poruszającego się po prostym odcinku szosy i rozpoczynającego hamowanie od szybkości 20m/s ma

Rasek [7]

Answer:

s = 22.5 m

Step-by-step explanation:

the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.

v (t) = 25 - 5 t

at t = 0 , v = 20 m/s

Let the distance is s.

$s =\int v(t) dt\\\\s =\int (25 - 5t)dt\\\\s= 25 t - 2.5 t^2 \\$

Let at t = t, the v = 20

So,

20 = 25 - 5 t

t = 1 s

So, s = 25 x 1 - 2.5 x 1 = 22.5 m

3 0

2 years ago

What’s the answer to this?

Elina [12.6K]

Step-by-step explanation:

Line is passing through the points:

$(-2,\:3)=(x_1,\:y_1) \:\&\: (2,\:5)=(x_2 ,\:y_2)$

Equation of line in two point form is given as:

$\frac{y -y_1 }{y_1 -y_2 } = \frac{x -x_1 }{x_1 -x_2 } \\ \\ \therefore \frac{y -3}{ 3 -5} = \frac{x -(-2) }{-2 -2} \\ \\ \therefore \frac{y -3}{ - 2} = \frac{x -(-2) }{-4} \\ \\ \therefore \frac{y - 3}{1} = \frac{x +2}{2} \\ \\ \therefore \: y-3= \frac{x}{2} + \frac{2}{2} \\ \\\therefore \: y= \frac{x}{2} + 1 +3\\ \\ \huge \red{ \boxed{\therefore \: y= \frac{1}{2} \:x+ 4}}\\ is \: the \: required \: equation \: of \: line.$

4 0

3 years ago

How do u factor this

PilotLPTM [1.2K]

You can factor a parabola by finding its roots: if

$p(x)=x^2+bx+c$

has roots $x_1,\ x_2$ , then you have the following factorization:

$p(x) = (x-x_1)(x-x_2)$

In order to find the roots, you can use the usual formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

In the first example, this formula leads to

$x_{1,2} = \dfrac{-2\pm\sqrt{4+4}}{2} = \dfrac{-2\pm\sqrt{8}}{2} = \dfrac{-2\pm2\sqrt{2}}{2} = 1 \pm \sqrt{2}$

So, you can factor

$x^2-2x-1 = (x-1-\sqrt{2})(x-1+\sqrt{2})$

The same goes for the second parabola.

As for the third exercise, simply plug the values asking

$f(1.5)=-5.25$

you get

$f(-1.5) = 1.5c-3 = -5.25$

Add 3 to both sides:

$1.5c = -2.25$

Divide both sides by 1.5:

$c = 1.5$

7 0

3 years ago

When 3^2 – 2 + 7 is subtracted from 5^3 – 4^2 + 10, what is the difference?

xxTIMURxx [149]

The difference is 105
3^2 - 2 + 7 =
(3 x 3) - 2 + 7 =
9 - 2 + 7 = 14

5^3 - 4^2 + 10 =
( 5 x 5 x 5) - ( 4 x 4) + 10
125 - 16 + 10 = 119

119 - 14 = 105

6 0

3 years ago