The answer for this question is 4.
The diagonals of a parallelogram bisect each other.
DH = HF and GH = HE
x + 5 = 2y
3x - 1 = 5y + 4
Solve the first equation for x.
x = 2y - 5
Now substitute 2y - 5 for x in the second equation.
3(2y - 5) - 1 = 5y + 4
6y - 15 - 1 = 5y + 4
6y - 16 = 5y + 4
y = 20
Now substitute 20 for y in the first original equation.
x + 5 = 2y
x + 5 = 2(20)
x + 5 = 40
x = 35
Answer: x = 35 and y = 20
Step One - List the factors of 60.
Step Two - Locate the factors that are seven apart from each other.
Factors of 60:
1 × 60
2 × 30
3 × 20
4 × 15
5 × 12
6 × 10
60 - 1 = 59
30 - 2 = 28
20 - 3 = 17
15 - 4 = 11
12 - 5 = 7
10 - 6 = 4
5 is 7 less than 12, and 60 is their least common multiple.
Answer: 5 and 12
Sadly, after giving all the necessary data, you forgot to ask the question.
Here are some general considerations that jump out when we play with
that data:
<em>For the first object:</em>
The object's weight is (mass) x (gravity) = 2 x 9.8 = 19.6 newtons
The force needed to lift it at a steady speed is 19.6 newtons.
The potential energy it gains every time it rises 1 meter is 19.6 joules.
If it's rising at 2 meters per second, then it's gaining 39.2 joules of
potential energy per second.
The machine that's lifting it is providing 39.2 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(2)(4) = 4 joules.
<em>For the second object:</em>
The object's weight is (mass) x (gravity) = 4 x 9.8 = 39.2 newtons
The force needed to lift it at a steady speed is 39.2 newtons.
The potential energy it gains every time it rises 1 meter is 39.2 joules.
If it's rising at 3 meters per second, then it's gaining 117.6 joules of
potential energy per second.
The machine that's lifting it is providing 117.6 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(4)(9) = 18 joules.
If you go back and find out what the question is, there's a good chance that
you might find the answer here, or something that can lead you to it.
Answer:
Unit rate is found simply by dividing.
6 pages of instruction/2 pages of practice problems = 3 pages of instruction per page of practice problems
Step-by-step explanation:
