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klemol [59]
3 years ago
5

What is the product of 2p + q and –3q – 6p + 1?

Mathematics
1 answer:
otez555 [7]3 years ago
5 0
The answer is -4p-2q+1
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A hamster runs at 17 centimeters per second in a wheel of radius 15 centimeters how fast will the wheel spin in revolutions per
dolphi86 [110]

Answer:

10.82 revolutions per mintue

Step-by-step explanation:

First we need the circumference of the cirkel. As you may know the formula for that is 2×pi×r, with r as the radius of the cirkel. So 2×pi×15=30pi centimeters or 94.24 centimeters. Now we will calculate the fraction of 17 centimeters on 94.24 centimeters. 17/94.24=0.18. That is how fast the wheel will spin. To get it in minutes we now do: 0.18×60=10.82 revolutions per minute.

4 0
3 years ago
Please helo i will make brainless for correct answers!​
elena-s [515]
False is the correct answer
7 0
4 years ago
Read 2 more answers
Maria brought home a sandwich that was 6 1/8 feet long to share with 4 members of her family. If each person, including Maria, a
mamaluj [8]
There are 5 people. Each ate the same amount. So it is 1/5 of the sandwich. 6.125/5=1.225 feet
6 0
3 years ago
0.8/y =2: 3/4 Please help
bulgar [2K]

Answer:

0.3

Step-by-step explanation:

0.8/y = 2 : 3/4

The value of y in the above expression can be obtained as follow:

0.8/y = 2 : 3/4

0.8/y = 2 ÷ 3/4

0.8/y = 2 × 4/3

0.8/y = 8/3

Cross multiply

0.8 × 3 = y × 8

2.4 = 8y

Divide both side by the coefficient of y i.e 8

y = 2.4/8

y = 0.3

Thus, the value of y is 0.3

8 0
3 years ago
The large piston in a hydraulic lift has an area of 2m^2. What force must be applied to the small piston with an area of .2m^2 i
Helen [10]

Answer:

147,000N

Step-by-step explanation:

A1= 2m^2

A2= 0.2m^2

F2= 14,700N

Required

F1, the applied force

Applying the formula

F1/A1= F2/A2

substute

F1/2=14700/0.2

2*14700= F1*0.2

29400= F1*0.2

F1= 29400/0.2

F1=147,000N

Hence, the applied force is 147,000N

3 0
3 years ago
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