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snow_lady [41]
3 years ago
13

50, 40 What’s the GCF of these numbers

Mathematics
2 answers:
ella [17]3 years ago
5 0

Answer: 10

Step-by-step explanation: To find the greatest common factor or <em>GCF</em> of 50 and 40, we begin by finding all of the factors of each number.

To find the factors of 50, we know that 50 ÷ 1 is 50 so 1 and 50 are factors.

50 ÷ 2 is 25 so 2 and 25 are factors.

We can't divide 50 evenly by 3 or 4 so we know they aren't factors.

50 ÷ 5 = 10 so 5 and 10 are factors.

However, if we continue to divide 50 by 6, 7, 8, and so on, we won't find any new factors. So the factors of 50 are 1, 2, 5, 10, 25, and 50.

Now let's find the factors of 40.

40 ÷ 1 is 40 so 1 and 40 are factors.

40 ÷ 2 is 20 so 2 and 20 are factors.

We can't divide 40 evenly by 3.

40 ÷ 4 is 10 so 4 and 10 are factors.

40 ÷ 5 is 8 so 5 and 8 are factors.

However, if we continue to divide by 6, 7, 8, and so on, we won't find any new factors.

So the factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40

Now that we have our list of factors, to find the greatest common factor, we simply find the largest number that is shared by the two lists.

Factors of 50 - 1, 2, 5, 10, 25, and 50

Factors of 40 - 1, 2, 4, 5, 8, 10, 20, and 40

Notice that both lists have a 50 as a factor and there is nothing larger than 50 that appears in both lists. So the greatest common factor of 50 and 40 is 10.

dexar [7]3 years ago
3 0

Answer:

10

Step-by-step explanation:

40=2 x 2 x 2 x 5

50=2 x 5 x 5

therefore, GFC = 2x5

GFC= 10

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The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

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Initial speed of debris is, s=120\ ft/s

The height 'h' of the debris above the ground is given as:

h(t)=-16t^2+120t

As per question, h(t)=56\ ft. Therefore,

56=-16t^2+120t

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0

Using quadratic formula to solve for 't', we get:

t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s

Therefore, the debris will reach a height of 56 ft twice.

When time t=0.5\ s during the upward journey, the debris is at height of 56 ft.

Again after reaching maximum height, the debris falls back and at t=7\ s, the height is 56 ft.

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