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2 years ago

Select all ratios equivalent to 30:6

Mathematics

1 answer:

Korolek [52]2 years ago

3 0

Answer:

60 : 12

90 : 18

120 : 24

150 : 30

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a stack of congruent blocks, piled one on top of the other, is 4/5 foot high. Each block is 1/10 foot tall. How many blocks are

kvasek [131]

8 blocks because 2/10=1/5 so if they total 4/5 then 2/10+2/10+2/10+2/10=8/10=4/5

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3 years ago

olya-2409 [2.1K]

The answer is: Perpendicular, because both equations intersects or cross over each other. There is an attached photo

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3 years ago

Read 2 more answers

Loretta and her family are going on vacation. Their destination is 610 miles from their home. Loretta is going to share some of

alexandr402 [8]

Loretta:
Average driving speed = 55 mph
Time = 4 hours
Distance traveled = 55*4 = 220 miles

Dad:
Average driving speed = 65 mph
Let x = hours driven
Distance traveled = 65x miles

The length of the trip is 610 miles, therefore
65x + 220 = 610
65x = 390
x = 6 hours

Answer: Loretta's dad drove for 6 hours.

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3 years ago

11. What is the name of Angle 7 and Angle 8?

Sever21 [200]

Answer:

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8 0

3 years ago

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg.

muminat

Answer:

Trial- 2 shows the conservation of momentum in a closed system.

Step-by-step explanation:

Given: Mass of balls are $m= 1.0\ kg$

Conservation of momentum in a closed system occurs when momentum before collision is equal to momentum after collision.

Let initial velocity of ball $A\ is\ u_1$
Initial velocity of ball $B\ is\ u_2$
Final velocity of ball $A\ is\ v_1$
Final velocity of ball $B\ is\ v_2$
Momentum before collision $= mu_1+mu_2$
Momentum after collision $=mv_1+mv_2$

Now, According to conservation of momentum.

Momentum before collision = Momentum after collision

$mu_1+mu_2=mv_1+mv_2$

We will plug each trial to this equation.

Trial 1

$mu_1+mu_2=mv_1+mv_2\\1.0(1)+1.0(-2)=1.0(-2)+1.0(-1)\\1-2=-2-1\\-1=-3$

Trial 2

$mu_1+mu_2=mv_1+mv_2\\1.0(.5)+1.0(-1.5)=1.0(-.5)+1.0(-\.5)\\.5-1.5=-.5-.5\\-1=-1$

Trial 3

$mu_1+mu_2=mv_1+mv_2\\1.0(2)+1.0(1)=1.0(1)+1.0(-2)\\2+1=1-2\\3=-1$

Trial 4

$mu_1+mu_2=mv_1+mv_2\\1.0(.5)+1.0(-1)=1.0(1.5)+1.0(-1.5)\\.5-1=1.5-1.5\\-.5=0$

We can see only Trial 2 satisfies the princple of conservation of momentum. That is momentum before collison should equal to momentum after collision.

5 0

3 years ago