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3 years ago

What is the GREATEST COMMON FACTOR of 20x2 , 10x , and 15?

Mathematics

2 answers:

Umnica [9.8K]3 years ago

7 0

Answer:

Step-by-step explanation:

when you divide all the numbers into 5 this is the most greatest common factor

irina [24]3 years ago

6 0

Answer:

Step-by-step explanation:

x can only go into 20x2 and 10x and 5 is the Gcf of these three numbers

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The expression on the left side of an equation is shown below.

horsena [70]

Answer:

Step-by-step explanation: hhhhhhhhhhhhhhhhhhhhhhhhhh

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3 years ago

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Which set of transformations would prove ΔLMN ~ ΔPQR?

weeeeeb [17]

Answer:

dilation

Step-by-step explanation:

Well, you didn't give an option to choose from. I know that the thing in the middle is a similar symbol. Dilations can prove that both shapes are similar. If LMN dilates to a smaller shape PQR then both shapes are SIMILLAR.

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3 years ago

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I run a book club with n people, not including myself. Every day, for 365 days, I invite three members in the club to review a b

Bezzdna [24]

<h3>Answer: 15</h3>

========================================================

Explanation:

The order doesn't matter. A group like {A,B,C} is the same as {B,A,C}.

All that matters is the overall group rather than the positioning of the members.

We'll use the nCr combination formula since order doesn't matter.

The value of n is unknown, but we know that r = 3 members are to be selected.

Let's pick a value for n at random. Let's say n = 10.

Plug n = 10 and r = 3 into the nCr formula below.

$n C r = \frac{n!}{r!(n-r)!}\\\\10 C 3 = \frac{10!}{3!*(10-3)!}\\\\10 C 3 = \frac{10!}{3!*7!}\\\\10 C 3 = \frac{10*9*8*7!}{3!*7!}\\\\ 10 C 3 = \frac{10*9*8}{3!}\\\\ 10 C 3 = \frac{10*9*8}{3*2*1}\\\\ 10 C 3 = \frac{720}{6}\\\\ 10 C 3 = 120\\\\$

Unfortunately we don't reach 365 or larger.

Let's try n = 11

$n C r = \frac{n!}{r!(n-r)!}\\\\11 C 3 = \frac{11!}{3!*(11-3)!}\\\\11 C 3 = \frac{11!}{3!*8!}\\\\11 C 3 = \frac{11*10*9*8!}{3!*8!}\\\\ 11 C 3 = \frac{11*10*9}{3!}\\\\ 11 C 3 = \frac{11*10*9}{3*2*1}\\\\ 11 C 3 = \frac{990}{6}\\\\ 11 C 3 = 165\\\\$

We're still under our target. The good news is that the nCr value is increasing.

So the idea is to do trial and error with various values of n. Keep incrementing n until nCr = nC3 is equal to 365 or larger.

Here's a table of values where r = 3 the entire time

$\begin{array}{|c|c|} \cline{1-2}\text{n} & \text{nCr}\\\cline{1-2}10 & 120\\\cline{1-2}11 & 165\\\cline{1-2}12 & 220\\\cline{1-2}13 & 286\\\cline{1-2}14 & 364\\\cline{1-2}15 & 455\\\cline{1-2}\end{array}$

The nCr values are also found in Pascal's Triangle. Each of those values are the fourth entry of each row.

When n = 14, we have nCr = 364 which is very close. We're one short unfortunately.

So we have to go for n = 15 instead. This makes the nCr value well over 365 of course, but it guarantees that you'll have plenty of trios to choose from such that no group of three is repeated. Unfortunately some trios will be left out.

7 0

2 years ago

Henry wanted to collect 65 pounds of cans in 7 days. How much should he collect each day to reach his goal? Which two whole numb

Marysya12 [62]

Answer:10 cans per day
Divide 65 by 7 and increase by 1 so that you get a number higher and not less than 65

7 0

3 years ago

Angle M and angle R are supplementary.

Burka [1]

Answer:

R = 150°

Step-by-step explanation: