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3 years ago

Find US, find MN, find YZ

Mathematics

1 answer:

uranmaximum [27]3 years ago

6 0

Answer:

I think the answer it 5

Step-by-step explanation:

You might be interested in

A circle has a diameter with endpoints at (6, 5) and (8, 5). Write the equation for the circle.

mario62 [17]

Answer:

$(x-7)^2+(y-5)^2=1$

Step-by-step explanation:

The two things that are required to formulate the equation of the circle is the center coordinate and the radius of the circle!

Center of the circle:

The center of the circle always lies at the midpoint of the endpoints of its diameter: Let's call the endpoints A(6,5) and B(8,5).

Using the midpoint formula we'll get:

$(x_m, y_m) = \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$(x_m, y_m) = \left(\dfrac{6+8}{2},\dfrac{5+5}{2}\right)$

$(x_m, y_m) = (7,5)$

This is the center coordinate of our circle.

Radius:

The radius of the circle is the distance from the center of the circle to any of the endpoints of the diameter (A or B)

We can use the distance formula:

$r = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$r = \sqrt{(x_1-x_m)^2+(y_1-y_m)^2}$

$r = \sqrt{(6-7)^2+(5-5)^2}$

$r = \sqrt{1^2}$

$r = 1$

Equation of the circle:

The equation is written as:

$(x-a)^2+(y-b)^2=r^2$

here, (a,b) are the center points of the circle

in our case this is $(a,b)=(x_m,y_m)=(7,5)$

and r = 1

$(x-7)^2+(y-5)^2=1^2$

$(x-7)^2+(y-5)^2=1$

This is the equation of the circle!

3 0

3 years ago

Aye help me please :)

mina [271]

Answer:

there's no solution.

Step-by-step explanation:

if y = 3x -4 then use this instead of y in the second equation

9x -3(3x+4) = 14 ➡9x-9x -12 = 14 9x will eliminate -9x and -12 can't be = 14 so there's no solution

4 0

3 years ago

"let v = r 2 with the usual addition and scalar multiplication defined by k(u1, u2) = (ku1, 0). determine which of the five axio

expeople1 [14]

Let $\mathbf u\in\mathbb R^2$ , where

$\mathbf u=(u_1,u_2)$

and let $k\in\mathbb R$ be any real constant.

Given this definition of scalar multiplication, we can see right away that there is no identity element $e$ such that

$e\mathbf u=\mathbf u$

because

$e\mathbf u=e(u_1,u_2)=(eu_1,0)\neq(u_1,u_2)=\mathbf u$

5 0

3 years ago

NEED HELP ASAP PLEASE!

Ivan

Answer:

i think the answer is b

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

It's about ratios I'm just kinda stuck

Blababa [14]

Answer:

Step-by-step explanation:

These answers are in order btw!

1=2

2=4

4=6

8=10

Table 2:

1=3

2=6

3=9

4=12

5=15

I’ll finish table 3 and put it in the commenta since this one will take me longer

8 0

3 years ago

Read 2 more answers