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Sever21 [200]
3 years ago
14

Joe ran 3 miles yesterday and wants to run at least 15 miles

Mathematics
1 answer:
igomit [66]3 years ago
4 0

Answer: 3(x)≥15 →x≥5

Step-by-step explanation:

Set it up with x as the number of days. Then multiply that by 3, so that you get 3(x). This will show that for an x amount of days, you ran 3 miles. then set it to MORE THAN or EQUAL TO 15. Then divide both sides by 3, and you get the above answer.

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Use the graphs of f and g to solve Exercises 87, 88, and 89.
Nadya [2.5K]

Answers and explanations:

87. The domain of added functions includes the restrictions of both. So the range of the added function in this question is [-4, 3]

88. When finding the domain of a divided function we do the same as adding, but with an extra rule: g can't equal zero. So for this question the domain is (-4, 3)

89. To graph f + g you add the y-values for each x-value. I added a picture to help explain this one!

4 0
3 years ago
1948 Olympics
mash [69]

Answer:

For 1948 Men's :Interquartile range is 1.5, Median is 58.3

For 2012 Men's: Interquartile range is 0.315, Median is 47.86

You can infer that in 2012 the swimmers were better and it was more competitive as the interquartile range was lower and the median was also lower as well

Step-by-step explanation:

1948 Men's 100m

57.3, 57.8, 58.1, 58.3, 58.3, 59.3, 59.6,1:00.5

               ⬆                ⬆                 ⬆

         Low IQR      Median       High IQR

Low IQR is average of 57.8 and 58.1 = 57.95

High IQR  is average of 59.3 and 59.6 =  59.45

Median is average of 58.3 and 58.3 = 58.3

Interquartile range is High IQR- Low IQR

59.45 - 57.95=1.5

2012 Men's 100m

47.52, 47.53, 47.8, 47.84, 47.88, 47.92, 48.04, 48.44

                    ⬆                  ⬆                      ⬆

             Low IQR         Median          High IQR

Low IQR is average of 47.53 and 47.8 = 47.665

High IQR is average of 48.04 and 47.92 = 47.98

Median is average of 47.88 and 47.84 = 47.86

Interquartile range is High IQR-Low IQR

47.98-47.665=0.315

Read more on Brainly.com - brainly.com/question/14915771#readmore

8 0
4 years ago
I need help on this please
12345 [234]
She is so right Because said it bro
6 0
2 years ago
Express sin A,cos A and tan A as ratios
OverLord2011 [107]

Answer:

Part A) sin(A)=\frac{2\sqrt{42}}{23}

Part B) cos(A)=\frac{19}{23}

Part C) tan(A)=\frac{2\sqrt{42}}{19}

Step-by-step explanation:

Part A) we know that

In the right triangle ABC of the figure the sine of angle A is equal to divide the opposite side angle A by the hypotenuse

so

sin(A)=\frac{BC}{AB}

substitute the values

sin(A)=\frac{2\sqrt{42}}{23}

Part B) we know that

In the right triangle ABC of the figure the cosine of angle A is equal to divide the adjacent side angle A by the hypotenuse

so

cos(A)=\frac{AC}{AB}

substitute the values

cos(A)=\frac{19}{23}

Part C) we know that

In the right triangle ABC of the figure the tangent of angle A is equal to divide the opposite side angle A by the adjacent side angle A

so

tan(A)=\frac{BC}{AC}

substitute the values

tan(A)=\frac{2\sqrt{42}}{19}

8 0
3 years ago
Hank’s pickup can travel 60 miles on 5 gallons of gas. How many gallons will Hank’s pickup need to travel 36 miles
Scorpion4ik [409]

60 / 5 = 12 miles per gallon


36 miles = 12 miles / gal *  x gallons

divide  by 12

36/12 =x gallons

3  =x

He needs 3 gallons


7 0
3 years ago
Read 2 more answers
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