Helpp me in geometry please

Mathematics

1 answer:

LekaFEV [45]2 years ago

4 0

Hi there !

You have a formule. You just have to apply it.

Let's do it then !

First of all, you must write the coordinates of the point E and F.

Therefore :

$E(-2;-4)$
$F(2;2)$

So here, we have :

$x_{1} =-2$
$y_{1} =-4$

$x_{2} =2$
$y_{2} =2$

Now, Let's apply the formula :

$M((-2+2)/2 ; (-4+2)/2)$
$M(0;-1)$

In short, the answer would be : B) $(0;-1)$ .

Hope this helps !

Photon

You might be interested in

The product of two integers is 72. One number is two less than five times the other.

Mama L [17]

Other number: x
one number: 5x - 2

x (5x - 2)=72
5x^2 - 2x = 72

7 0

2 years ago

Read 2 more answers

Sketch the region enclosed by y=3x and y=x^2 . Then find the area of the region.

kolezko [41]

Answer:

$A=4.5u^2$

Step-by-step explanation:

The integral of a function gives you the area under the curve, the subtraction of one of the areas from the other will give you the area in between.

The limits of integration are the points where the curves intersect each other(take the curves has a system of equation and solve for x and y):

$y=3x, y=x^2\\3x=x^2\\x^2-3x=0\\x(x-3)=0\\x_1=0\\x-3=0\\x_2=3$

$y_1=3x_1=3(0)=0\\y_2=3x_2=3(3)=9$

The integral will be the subtraction of the curve $y=3x$ and $y=x^2$ (In the graph you can see y=3x is the upper curve):

$\int\limits^3_0 {3x-x^2} \, dx =\left\frac{3x^2}{2}-\frac{x^3}{3}\right|^3_0=\frac{3(3)^2}{2}-\frac{(3)^3}{3}-\left(\frac{3(0)^2}{2}-\frac{(0)^3}{3}\right)\\\int\limits^3_0 {3x-x^2} \, dx =\frac{27}{2}-\frac{27}{3} =13.5-9=4.5u^2$