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Anarel [89]
3 years ago
9

Can someone help me please ASAP?

Mathematics
1 answer:
Mariana [72]3 years ago
5 0
Left is greater.

1 > -1
You might be interested in
Is (6,-10) a solution of y=3x-8
Vitek1552 [10]
You can easily test this if you know that (6, -10) corresponds to (X, Y). Knowing this, you can: 

X = 6
Y = -10

you put this into your equation:

-10 = 3*6 - 8

calculate it:

-10 = 18 - 8
-10 = 10
This is not true of course, -10 is not equal to 10. Therefore, (6, -10) is not a solution of y = 3x-8 :)
6 0
3 years ago
I am a 2 digit number.
AlexFokin [52]

Answer:

65

Step-by-step explanation:

6+5=11

5 0
3 years ago
How do I do this?? 7-11
mihalych1998 [28]

Answer: slope intercept is y=mx+b form s b= the y intercept and x will just be x and the m is usually a fraction which tells you the slope so what you need to do is mark these points on a line graph draw a line from one to the other and every where it crosses over a point is where you can find the slope and the y intercept giving you the answer



7 0
3 years ago
Lisa drove 7,000 miles in 70 days. She drove the same number of miles each day.
inysia [295]

Answer: I'd assume its B 100 miles each day.

Step-by-step explanation:

Because 70 x 100 is 7,000 which is equal to the amount of miles stared in the question

5 0
3 years ago
Read 2 more answers
What is the length of the segment connecting the points A(1,3) and B(6,5)?
Natalija [7]

Answer:

The answer is

<h2>\sqrt{29}  \:  \:  \: or \:  \:  \:5.38 \:  \:  \: units</h2>

Step-by-step explanation:

The length of the segment connecting two points can be found by using the formula

d =  \sqrt{ ({x1 - x2})^{2} +  ({y1 - y2})^{2}  }  \\

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

A(1,3) and B(6,5)

The length is

|AB|  =  \sqrt{ ({1 - 6})^{2} +  ({3 - 5})^{2}  }  \\  =  \sqrt{ ({ - 5})^{2}  + ( { - 2})^{2} }  \\  =  \sqrt{25 + 4}  \\  =  \sqrt{29}

We have the final answer as

\sqrt{29}  \:  \:  \: or \:  \:  \:5.38 \:  \:  \: units

Hope this helps you

3 0
2 years ago
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