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Korolek [52]
4 years ago
9

Starting at the same point, Joe and Anita go roller blading in opposite directions. If Joe roller blades at a speed of 4 mph, an

d Anita roller blades at a speed of 4 mph, how far apart will they be in 2 hours?
Mathematics
1 answer:
agasfer [191]4 years ago
8 0

Answer:

They will be at a distance of 16 miles apart from each other.

Step-by-step explanation:

Both Joe and Anita start roller balding in opposite directions, and both of them have the same speed of 4 mph.

As they move oppositely in a straight line therefore, the distance between them will be the distance travelled by each of them's sum.

Distance covered by Joe in 2 hours= 8 miles

Distance covered by Anita in 2 hours= 8 miles

Therefore, the total distance from each other =16 miles

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Evaluate b/a−1.2 when a=4 and b=12
Elena L [17]

Answer:

3-1.2=1.8

Step-by-step explanation:

12/4=3-1.2=1.8

3 0
3 years ago
4. If A. 35<br> B.180<br> C.125<br> D.55
Flura [38]
Its B. 180 trust lol i got you
8 0
3 years ago
Read 2 more answers
Mrs. Nestler enjoys listening to classical music. She has the following audio CDs by her favorite composers in her collection: 4
ivolga24 [154]

Answer:

10.55% probability

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the CDs are chosen is not important. So we use the combinations formula to solve this question.

1 Bach CD, from a set of 4.

1 Beethoven CD, from a set of 6.

1 Brahms CD, from a set of 3.

1 Handel CD, from a set of 2.

So, D=144

4 CDs from a set of 4+6+3+2 = 15.

So, T= 1365

p= D/T= 144/1365 = 0.1055

10.55% probability that she will choose one by each composer

4 0
3 years ago
Ill give you brainliest pleaseeeeeeee help me
ruslelena [56]
Line D and C because they are not parallel but distorted or unsymmetrical
4 0
4 years ago
Help please! (no one was answering it in physics and i really need help)
aivan3 [116]

Let v_0 be the cat's speed just as it leaves the edge of the table. Then taking the point 1.3 m below the edge of the table to be the origin, the cat's horizontal position at time t is given by

x(t)=v_0t

and its height is

y(t)=1.3\,\mathrm m-gt^2

where g is 9.8 m/s^2, the magnitude of the acceleration due to gravity.

The time it takes for the cat to hit the ground is t with

0=1.3\,\mathrm m-gt^2\implies t=\sqrt{\dfrac{1.3\,\rm m}g}\approx0.36\,\mathrm s

(Unfortunately, this doesn't match any of the given options...)

The cat lands 0.75 m away (horizontally) from the edge of the table, so that its speed v_0 was

0.75\,\mathrm m=v_0(0.36\,\mathrm s)\implies v_0\approx2.08\dfrac{\rm m}{\rm s}

(Again, not one of the answer choices...)

I'm guessing there's either a typo in the question or answers.

6 0
3 years ago
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