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ELEN [110]
3 years ago
7

The average speed to go by car from home to college is 30 miles per hour. The average speed to go back home from college is 20 m

iles per hour because of traffic. If the college is 60 miles from home, what is the average speed for the entire round trip? What is the average speed if the college is 40 miles from home?
Mathematics
1 answer:
Levart [38]3 years ago
3 0

Answer:

24 mph for both 60 miles and 40 miles.

Step-by-step explanation:

We need to find how long it takes to get to college and from it.

Speed x time (x) = Distance (60)

30x = 60

x=2

It takes two hours to get to college

20x = 60

x = 3

It takes 3 hours to get back. It's a total of 5 hours, 2 hours at 30mph and 3 at 20. Now we need to find the average

30+30+20+20+20=120

120/5= 24

The average speed for the round trip is 24.

To do the second part, we follow the same process, but replace 60 with 40 for distance.

30x=40

x= 1.33

20x=40

x=2

We can add the total distance and divide by total time to find the average.

The total distance is 80. Time is 3.33

80/3.33=24

The average speed is still 24.

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sdas [7]

Answer:

Infinite solutions

Step-by-step explanation:

You can put in any number for x and the 2 sides of the equation will equal each other.

5 0
2 years ago
Read 2 more answers
write an algebraic expression for each verbal expression... 1. the sum of a number and ten 2. 15 less than k 3.the product of 18
erica [24]

Answer:

Step-by-step explanation:

N+10

K-15

18*2

6+(2*m)

3*n+8

(5*n)-17

2+(y^2)

(G^4)-9

7 0
3 years ago
Pleaseeee help on question 3 and 5 pleaseeeee wuick thank youuuuuuuuu
elena55 [62]

Answer:

3a, 13cm 5, 1:200

Step-by-step explanation:

In the question 3 you just have to divide all of those numbers with that 5000, maybe also make them cm so it makes more sense. So 650m = 65000cm:5000= 13cm

then for the question 5, make the km to cm, so 4.2km would be 420000cm. Then you have to divide that 420000cm with the 21 cm, so 420000:21= 200. Then the scale should be 1:200

I had difficulties with these as well. If you need extra help with the task 3 please pm me :)

7 0
3 years ago
Find the GCF of 4g^4 and 34g^3.
USPshnik [31]
Yes, find the prime factors.

4 can be written as 2*2.
g^4 can be written as g*g*g*g.

That is the prime factorization of the first expression.
4g^4 can be written as 2*2*g*g*g*g.

34 can be written as 2*17.
And again we can split up the g business: g^3 is g*g*g.
34g^3 can be written as 2*17*g*g*g.

What do these expressions have in common?
They have a 2 in common,
and three of the g's in common, ya?

Therefore the GCF(4g^4, 34g^3) = 2g^3.
6 0
3 years ago
Please help me do this iready please i am saying it kindly
abruzzese [7]

Answer:

See below.

Step-by-step explanation:

The answer is each figure in group 2 has one face.

-hope it helps

6 0
2 years ago
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