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oksian1 [2.3K]
3 years ago
9

In 3 seconds, kevin can dribble a basket ball 13 times. if kevin dribbles at this same rate for 27 seconds, how many times will

he dribble the basket ball?

Mathematics
1 answer:
Karolina [17]3 years ago
4 0
I hope this helps you

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What is |–93|?<br> i need help
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Answer:

93

Step-by-step explanation:

The absolute value of any number will always be positive. Therefore, it is automatically 93.

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In the diagram, polygon ABCD is flipped over a line of reflection to form a polygon with its vertices at A, B, and D are shown,
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The line of reflection is at 5 on the x-axis. C is at the point,(6,2)
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Four students are working in a group and are asked to find the length of BC. Which two students set up the problem correctly? Mo
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The best and the most correct answer among the choices provided by the question is the first choice. The two students that set up the problem correctly is <span> Moe: x = (5 - 1)2 + (1 - 4)2 and Jimmy: x2 = 32 + 42.</span> I hope my answer has come to your help. God bless and have a nice day ahead!
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The speed(S) of a car varies partly directly as its mass(M) and partly directly as the quantity (Q) of fuel in it. When the spee
weeeeeb [17]

Answer:

S varies partly directly as M and Q.

S=C.

S=KMQ+C.

For the first one...

speed=80,m=220,Q=30.

80=K20×30+C.

80=600K+C......(I).equation one.

For the second one....

speed=60,m=300,Q=40.

60=K300×40+C.

60=12000K+C.....(ii). equation two.

Minus eqtn(I) from eqtn(ii).

80=600K+C.

- 60=12000K+C.

K=0.01754~0.018.

Substitute K=0.018 into eqtn(I).

80=600K+C

80=600×0.018+C.

80=10.8+C.

C=80-10.8=69.2.

The relation is S=0.018MQ+69.2

when speed is 100 and mass is 250 find the volume.

100=0.018×250×Q+69.2.

100=4.5Q+69.2.

4.5Q=100-69.2

4.5Q=30.8.

Q=30.8/4.5.

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6 0
3 years ago
Half an hour after leaving her house Beverly was 20 miles from home. At this point she merged onto the highway and was traveling
N76 [4]

Answer:

1. 45.71 miles

2. 10 miles

3. 89.43 miles

4. 0.79 hr

5. 0.025 hr

Step-by-step explanation:

speed = distance/time

After 4 hrs Beverly was 200 miles from home , however she had spent 0.5hr travelling for 20 miles prior to getting to the highway.

Therefore;

Constant speed before Highway = 20/0.5 = 40 miles/hr

Constant speed at Highway = \frac{(200-20)}{(4-0.5)} = 51.43 miles/hr

1. in 1 hr beverly will have travelled;

⇒ (20 miles in 0.5 hr ) + (0.5 hr  at speed 51.43 miles/hr)

⇒ 20 + (0.5 x 51.43) = 45.71 miles

2. In 0.25 hr beverly had not reached the highway

⇒ distance = speed x time

⇒ distance = 40 x 0.25 = 10 miles

3. in 1.85 hrs beverly will have travvelld;

⇒ (20 miles in 0.5 hr ) + (1.35 hr at speed 51.43 miles/hr)

⇒ 20 + 69.43 = 89.43 miles

4. Time taken to get to 35 miles

⇒ (20 miles in 0.5 hr ) + (15 miles  at speed 51.43 miles/hr)

⇒ 0.5hr + (15/51.43) hr = 0.79 hr

5. Time taken to travel 1 mile

 ⇒ time = distance / speed = 1 mile  / 40 miles/hr

⇒ time = 0.025 hr

8 0
3 years ago
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