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GuDViN [60]
3 years ago
14

A number is equal to twice a smaller number plus 3. The same number is equal to twice the sum of the smaller number and 1. How m

any solutions are possible for this situation?
Infinitely many solutions exist because the two situations describe the same line.
Exactly one solution exists because the situation describes two lines that have different slopes and different y-intercepts.
No solutions exist because the situation describes two lines that have the same slope and different y-intercepts.
Exactly one solution exists because the situation describes two lines with different slopes and the same y-intercept.
Mathematics
2 answers:
Marizza181 [45]3 years ago
8 0

Answer:

No solutions exist because the situation describes two lines that have the same slope and different y-intercepts.

Step-by-step explanation:

Small number = S

Number = N

N = 2S + 3

N = 2(S+1)

N = N

2S + 3 = 2(S + 1)

2S + 3 = 2S + 2

-2S         -2S

3 = 2

No solutions

Therefore...

"No solutions exist because the situation describes two lines that have the same slope and different y-intercepts." is correct

Svetlanka [38]3 years ago
4 0
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2 years ago
Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤
MrMuchimi
The average rate of change of a function f(x) in an interval, a < x < b is given by
\frac{f(b) - f(a)}{b - a}

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by \frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by \frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by \frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by \frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by \frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by \frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0
3 0
3 years ago
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Find the measure of angle s?<br><br> a. 90°<br> b. 45°<br> c. 30°<br> d. 15°
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8 0
2 years ago
Hurry in need of help! Please
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\\ \sf\dashrightarrow \dfrac{x+20}{2}=3x

\\ \sf\dashrightarrow x+20=2(3x)

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\\ \sf\dashrightarrow x=4

3 0
3 years ago
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