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vodomira [7]
3 years ago
10

Johnny was able to drive an average of 29 miles per hour faster on his car after the traffic cleared. He drove 30 miles in traff

ic before it cleared and the drove another 176 miles. If the total trip took 6 hours, then what was his average speed in traffic?
Mathematics
1 answer:
Fantom [35]3 years ago
3 0

Answer:

15 mph

Step-by-step explanation:

Before traffic cleared:

distance = 30 miles

average speed = s

time = t

d = st

30 = st

After traffic cleared:

distance = 176 miles

average speed = s + 29

time = 6 - t

d = st

176 = (s + 29)(6 - t)

30 = st

s = 30/t

176 = (30/t + 29)(6 - t)

176t = (30 + 29t)(6 - t)

176t = 180 - 30t + 174t - 29t^2

29t^2 + 32t - 180 = 0

(29t + 90)(t - 2) = 0

t = -90/29 or t = 2

Time must be a positive number, so we discard the solution t = -90/29.

t = 2

30 = st

30 = s * 2

s = 15

Answer: 15 mph

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Fred has 230 animals at his zoo. Grace at 349 more animals in her zoo than Fred has. how many animals do they have altogether? T
WARRIOR [948]

Step-by-step explanation:

okay so im in 8th grade and i just worked the problem out and i got , 809 ,

first you add 230 + 349 to see how much she has in her farm

she has 579, then you add 579 because she has that many , and he has 230 so you add 579 +230 and tou get 809

work:

230

+349

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579

+230

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809

hope i helped!!

4 0
4 years ago
A pilot flew a 400 mile flight in 2.5 hours flying into the wind. Flying the same rate and with the same wind speed the return t
eduard

Answer: 20 mph

Explanation:

Speed is a physical quantity which is equal to the ratio between the distance covered (d) and the time taken (t):

v=\frac{d}{t}

In the first part of the problem, the plane flew a distance of d=400 mi in a time of t=2.5 h. The speed of the plane in this case was the difference between the proper speed of the plane, v, and the speed of the wind, w, since the plane flew opposite to the wind. So we can write:

v-w=\frac{400mi}{2.5h}=160 mph (1)

During the return trip, the plane flew with a speed (v+w), since the wind was on the tail, and it took 2 hours to cover the same distance:

v+w=\frac{400 mi}{2 h}=200 mph (2)

So we have two equations with two unknown variables. From (1), we get

v=160+w

Substituting into eq.(2)

(160+w)+w=200\\160+2w=200\\2w=40\\w=20 mph

So, the speed of the wind was 20 mph.

3 0
3 years ago
Read 2 more answers
Write your question here (Keep it simple and clear to get the best answer)4x³-5x is it odd or even function
Dvinal [7]

Answer:

odd

Step-by-step explanation:

To determine if odd/ even

• If f(x) = f(- x) then it is even

• If f(- x) = - f(x) then it is odd

Given

f(x) = 4x³ - 5x , then

f(- x) = 4(- x)³ - 5(- x) = - 4x³ + 5x

Since f(x) ≠ f(- x) then it is not even

- f(x) = - (4x³ - 5x) = - 4x³ + 5x

Since f(- x) = - f(x) , then 4x³ - 5x is odd

5 0
3 years ago
Which of the following expressions is not equivalent to -12x+36?
Tju [1.3M]
The answer I believe is -3x + 20-9x+16
7 0
3 years ago
We often deal with weighted​ means, in which different data values carry different weights in the calculation of the mean. For​
Veseljchak [2.6K]

Answer:

a) The student's overall average for the class is 87.97.

b) He would need a score above 100 to get an A, which means that he could not have scored high enough on the final exam to get an A in the class.

Step-by-step explanation:

Weighed average:

To solve this question, we find the student's weighed average, multiplying each of his grade by his weights.

Grades and weights:

Scored 82.5 on the midterm, worth 30%.

Scored 88.6 on the final exam, worth 30%.

Scored 91.6 on the homework, worth 40%.

a. On a 100-point scale, what is the student's overall average for the class?

Multiplying each grade by it's weight:

A = 82.5*0.3 + 88.6*0.3 + 91.6*0.4 = 87.97

The student's overall average for the class is 87.97.

b. The student was hoping to get an A in the class, which requires an overall score of 93.5 or higher. Could he have scored high enough on the final exam to get an A in the class?

Score of x on the final class, and verify that the average could be 93.5 or higher. So

A = 82.5*0.3 + 88.6*0.3 + 0.4x

A \geq 93.5

82.5*0.3 + 88.6*0.3 + 0.4x \geq 93.5

0.4x \geq 93.5 - (82.5*0.3 + 88.6*0.3)

x \geq \frac{93.5 - (82.5*0.3 + 88.6*0.3)}{0.4}

x \geq 105.425

He would need a score above 100 to get an A, which means that he could not have scored high enough on the final exam to get an A in the class.

3 0
3 years ago
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