All categories

2 years ago

I need help with problem number 3 if someone can help me with that

Mathematics

1 answer:

Flura [38]2 years ago

3 0

Plug in x into the equation. Take x= -2 for example. Your equation is 2x-5, so instead of x, you put -2.
2(-2)-5 = -9
Your table is as follows:
(-2,-9)
(-1,-7)
(0,-5)
(1,-3)

You might be interested in

5 Victor knows that 2 + 7 = 7+2 and 2•7= 7•2

Romashka [77]

Answer:2^7 ≠ 7^2

Step-by-step explanation:

2^7 = 2 * 2 * 2 * 2 * 2 * 2 * 2

The only factor is 2

7^2 = 7 * 7

The only factor is 7

Comparing the factors when written as a repeated multiplication,

2^7 and 7^2 have no common factor :

2^7 = 2 * 2 * 2 * 2 * 2 * 2 * 2 = 128

7^2 = 7 * 7 = 49

6 0

2 years ago

A line segment has endpoints (-5, 10) and (a,b) . If the midpoint of the segment is (13, -2), what is the absolute difference be

AlexFokin [52]

If the line segment point is D(-5, 10) and E(a,b) and the midpoint of the segment is F(13, -2) that mean
DE= 2*DF

You can directly find the distance of AC
Xdf= Xf-Xd= 13 - (-5)= 18
Ydf= Yf - Yd= -2 - 10= -12

Then add the distance of AB( which is 2*AC) to point D
Xe= a = Xd + 2*Xdf
a= -5 +2*18= 31

Ye= b = Yd + 2Yf
b= 10+ 2*-12= -14

absolute difference between a and b:
|b-a|= |-14-31|= 45

7 0

2 years ago

PLEASE HELP I DONT KNOW

riadik2000 [5.3K]

Answer:

d, diameter

Step-by-step explanation:

radius is half of a diameter, or the distance from the center of the circle to the circumference

circle is a... circle

center is...

5 0

2 years ago

Read 2 more answers

SV bisects ∠TSU. Complete the proof that UV ≅ TV

NNADVOKAT [17]

Is there a chart involved

5 0

2 years ago

The midpoint of \overline{\text{AB}}AB is M(-6, -4)M(−6,−4). If the coordinates of AA are (-4, -3)(−4,−3), what are the coordina

valentina_108 [34]

Answer:

The coordinates of B is (-8,-5).

Step-by-step explanation:

The midpoint of line AB is M. The coordinate of M is (-6,-4).

The coordinates of A is (-4,-3)

We need to find the mid point of B.

If M(x,y) is the midpoint of the coordinates (x₁,y₁) and (x₂,y₂). The mid point theorem is used as follows :

$M(x,y)=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})$

Let the mid point of B is (x₂,y₂). Put (x,y) = (-6,-4), (x₁,y₁) = (-4,-3).

$(-6,-4)=(\dfrac{-4+x_2}{2},\dfrac{-3+y_2}{2})\\\\\dfrac{-4+x_2}{2}=-6\ \text{and}\ \dfrac{-3+y_2}{2}=-4\\\\-4+x_2=-12\ \text{and}\ -3+y_2=-8\\\\x_2=-12+4\ \text{and}\ y_2=-8+3\\\\x_2=-8\ \text{and}\ y_2=-5$

So, the coordinates of B is (-8,-5).

6 0

2 years ago