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3 years ago

Find the gcf and lcm of 24 and 40

2 answers:

6 0

F you just want to know what is the greatest common factor of 24 and 40, it is 8.
gcf (24,40) = 24×40lcm(24,40)=96012024×40lcm(24,40)=960120= 8

Alex3 years ago

5 0

Answer:

gcf: 8

Lcm: 120

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Suppose 100 people all toss a hat into a box and then proceed to randomly pick out a hat. what is the expected number of people

tatuchka [14]

1 person because there is 1 in 100 chance that you get your hat back

4 0

3 years ago

Jamie went to the his 6 friends. For each friend, he spent $4.75 for a sandwich, $1.25 for a cold beverage, and $.56 for a piece

masha68 [24]

Answer:

Jamie spent $39.36 in total.

Step-by-step explanation:

First, add the money spent for the sandwich, beverage, and fruit together.

4.75 + 1.25 + .56 = 6.56

Then, multiply the amount spent for one friend by six to find the total.

6.56 x 6 =39.36

I hope this helped you!

8 0

4 years ago

Solve the equation 7h-5(3h-8) = 72

frosja888 [35]

Answer:

h = -4

Step-by-step explanation:

7h-5(3h-8) = 72

Distribute

7h - 15h +40 = 72

Combine like terms

-8h +40 = 72

Subtract 40 from each side

-8h+40-40 = 72-40

-8h = 32

Divide each side by -8

-8h/-8 = -32/-8

h = -4

3 0

3 years ago

Read 2 more answers

Factor 6x^2 + 7x - 3

posledela

The correct answer would be (3x-1)(2x+3)

8 0

4 years ago

Read 2 more answers

A car has mass 1500 kg and is traveling at a speed of 35 miles/hour. what is its kinetic energy in joules? (Be sure to convert m

photoshop1234 [79]

Answer:

Factor by which kinetic energy increase = 4 times

Step-by-step explanation:

Given,

Mass of the car, v1 = 1500 kg

initial speed of car = 35 miles/h

$=\dfrac{35\times 1609.34}{3600}\ m/s$

= 15.64 m/s

Initial kinetic energy of the car is given by,

$k_1\ =\ \dfrac{1}{2}.m.v_1^2$

$=\ \dfrac{1}{2}\times 1500\times (15.64)^2\ joule$

= 183606.46 J

Final velocity of car v2 = 70 miles/hour

$=\dfrac{70\times 1609.34}{3600}$

= 31.29 m/s

So, final kinetic energy of car is given by

$k_2\ =\ \dfrac{1}{2}.m.v_2^2$

$=\ \dfrac{1}{2}\times 1500\times (31.29)^2$

= 734425.84 J

Now, the ratio of final to initial kinetic energy can be given by,

$\dfrac{k_2}{k_1}=\ \dfrac{734425.84}{183606.46}$

$=>\ k_2\ =\ 4k_1$

Hence, the kinetic energy will increase by 4 times.

7 0

3 years ago