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Karolina [17]
3 years ago
6

If you were to measure a person once a year, every year, from the ages of ten to twenty-five, would your ordered pairs of age an

d helght represent a function? по O yes​
Mathematics
1 answer:
Naily [24]3 years ago
4 0
No they wouldn’t because height dosent work at a steady pace
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2 groups of 3 tenths pls answer today I need this answer
SSSSS [86.1K]

Answer:

6/10

Step-by-step explanation:

2 groups of 3/10 =

2 × 3/10 =

6/10

3 0
3 years ago
(2x > 10) list some values for x that would make this inequality true
Aneli [31]

Answer:

6 , 7 , 8 , 9...

Step-by-step explanation:

2x > 10

x > 5

then

x = 6 , 7 , 8, 9...

I think that I am right and also think this will helpful to you.....

6 0
2 years ago
A girl has 98 beads, and all but 14 were lost. how many beads did she loose?<br>​
castortr0y [4]

Answer:

84 beads

Step-by-step explanation:

She had 98 beads and lost all but fourteen. So it would be 98 - 14 which would get you 84 beads that the girl has lost

8 0
3 years ago
I need help with this as soon as possible!​
Kay [80]

Answer:

1. -3/6 is negative, because it descends by 3 and goes over by 6.

2. My first answer literally gives the keyword negative, therefore it's a negative line because it has a <em>negative slope</em>.

1. -8/0 is undefined, because there is a rise but there is no run, which also means that it's a:

2. vertical line

1. -5/-10 simplifies to 5/10 or 1/2, which is positive because it rises by 5(or 1) and runs over by 10(or 2).

2. Thus, this means that it's a positive line since it has a <em>positive slope</em>.

1. 0/-2 is zero, because when you actually solve the equation it'll equal 0.

2. This means that it'll have a horizontal line since there is no rise but there is a run only.

8 0
2 years ago
To solve for y in the equation 2y + y = 5, what first step would need to be taken? Subtract 2 from each side. Add 2y and y. Subt
cricket20 [7]

Answer:

Add 2y and y.

Step-by-step explanation:

you have to combine what is common

plz give brainliest

4 0
3 years ago
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