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Taya2010 [7]
2 years ago
5

After 3.5 hours, Peyton had traveled 161 miles. If she travels at a constant speed, how far will she have traveled in 4 hours

Mathematics
1 answer:
Flura [38]2 years ago
4 0

Answer:

1.3 more miles i think

Step-by-step e

3.5 hours equal 210 and i divided it by 163 and thats what came up

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Which function is increasing at the highest rate? A. B. A linear function on a coordinate plane passes through (minus 1, 3), and
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The highest rate on increasing is of 12x -6y = -24; Option B is the correct answer.

The options are given in the image attached with the answer

<h3>What is a Function ?</h3>

A function is a law that relates a dependent variable and an independent variable.

It is asked among the options given , which is increasing at the highest rate.

To increase at a rate , the value of the slope , m should be > 0

For the Option 1

for a straight line function , the slope is given by

m = ( y₂ -y₁)/(x₂-x₁)

m = ( -3 -3)/(2- (-1)) = -6 /3 = -2

Therefore the function is decreasing

For Option 2

12x - 6y = -24

-6y = -12x -24

y = 2x +4

m = 2 (increasing)

For option 3

m = (-4 + 5)/(2-1) = 1

For Option 4

(8,0) (0,-4)

m = (-4 -0) /(0-8) = 4/8 = 1/2

The highest rate on increasing is of 12x -6y = -24

Therefore Option B is the correct answer.

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6 0
2 years ago
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Sixth sevenths times seven tenths
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Stefan won 54 lollipops playing hoops at the county fair. At school he gave three to every student in his math class. He only ha
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Compare the process of solving |x – 1| + 1 &lt; 15 to that of solving |x – 1| + 1 &gt; 15. Check all of the following you includ
zepelin [54]

The required Comparison of the inequalities are

  • The |x – 1| + 1 > 15 represents the value of x lies between 13<x<15.

The range of values encompassing the region's junction is (-13, 15).

  • If x is more than or equal to 15, then x-11+1>15 indicates the value of x is greater than or equal to 13. None of the regions in the intersection are empty.

<h3>What is inequality?</h3>

When comparing two numbers, an inequality indicates whether one is less than, larger than, or not equal to the other.

We take into account the various variables of the inequality

|x – 1| + 1 > 15

Therefore

|-x-1|+1-1<15-1

|-x-1|-1 <14

13<x<15

The required region lies between the inequality -13 <x< 15.

Simplify the inequality Ix-11+1 > 15 we get,

|x-1|+1 > 15

|x+1| +1-1 >15-1

|x-1| > 14

x> 15  

x<-13

  • If x has a value between -13 and x + 15, then the expression "|x-1|+1+115" is true. The range "(-13, 15)" contains the intersection of the region.
  • If "|x-1|+1>15" then either "x >15" or "x-13" applies to the value of x. This region's intersection is unoccupied.

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