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il63 [147K]
2 years ago
15

Select the ordered pair that is a solution to the function. (A point on the line.)?

Mathematics
1 answer:
soldier1979 [14.2K]2 years ago
6 0

Answer:

-7

Step-by-step explanation:

first you plug in f(3) in the x so -2(3)= -6

then -6 - 1 = -7

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15 Pts + Brainliest for the best answer
34kurt

If T is the midpoint of SU, then ST ≅ TU.

Therefore we have the equation:

6x = 2x + 32       <em>subtract 2x from both sides</em>

4x = 32       <em>divide both sides by 4</em>

x = 8

ST = 6x → ST = 6(8) = 48

TU = ST, therefore ST = 48

SU = ST + TU = 2ST, therefore SU = 2(48) = 96

<h3>Answer: ST = 48, TU = 48, SU = 96</h3>
3 0
3 years ago
Juan and Nina are in the Junior High Orchestra. As part of the orchestra they each have to learn two instruments. Juan is learni
larisa86 [58]

Answer:

Nina practiced the viola for 11.25 hours last week.

Step-by-step explanation:

Juan practiced the violin for 9 hours last week. For each 3 hours that he practices the violin, he practices 2.5 hours of cello. So

3h violin - 2.5h cello

9h violin - xh cello

3x = 9*2.5

x = 7.5

He practiced 7.5 hours of cello, the same as Nina.

For every 3 hours Nina spends practicing the viola she practices the cello for 2 hours. So:

3h viola - 2h cello

xh viola - 7.5h cello

2x = 3*7.5

x = 7.5*1.5

x = 11.25

Nina practiced the viola for 11.25 hours last week.

7 0
3 years ago
Pythagorean theorem
Talja [164]

Answer:

b=20 yards

Step-by-step explanation:

15^2+x^2=25^2

225+x^2=625

x^2=400

x=20

5 0
3 years ago
1. Jack is 30 years old and Alice is 35 years old. What percent of Alice's age is Jack, rounded to the nearest tenth of a percen
Firlakuza [10]
The answer would be 57percent
4 0
3 years ago
In a standard normal distribution, about _________% of the scores fall above a z-score of 3.00
Katarina [22]

The normal rule to remember is 68-95-99.7, i.e. plus or minus three sigma corresponds to 99.7% of the probability. That leaves 0.3% in the two tails, or 0.15% in the tail above 3 sigma.

Answer: 0.15%


7 0
3 years ago
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