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vlada-n [284]
4 years ago
14

Find the greatest common factor: 9a3bc2 + 12ac4 + 15a2b3c 3abc 15a3b3c4 3ac 9a2b2c2

Mathematics
1 answer:
VARVARA [1.3K]4 years ago
8 0
The greatest common factor is 3abc
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Why is it not possible to divide by zero
koban [17]

Answer:

In mathematics, division by zero is division where the divisor (denominator) is zero. Such a division can be formally expressed as a/0 where a is the dividend (numerator). In ordinary arithmetic, the expression has no meaning, as there is no number which, multiplied by 0, gives a (assuming a≠0), and so division by zero is undefined.

8 0
4 years ago
Read 2 more answers
Anyone solve for X?
Korvikt [17]
Since the triangles are congruent and m<I is congruent to m<P.
Therefore, 3x + 4 = 72 - x
3x + x = 72 - 4
4x = 68
x = 68/4 = 17
x = 17
8 0
4 years ago
Consider the data below. Find the change in the (a) mean, (b) median, and (c) mode when the
LuckyWell [14K]

Answer:

A

Step-by-step explanation:

3 0
3 years ago
if you draw 35 lines on a piece of paper so that no two lines are parallel to each other and no three lines are concurrent, how
Levart [38]

Answer:

The lines will intersect 595 times.

Step-by-step explanation:

We can do it in 2 methods.

<u>1st method:-</u> Let us count the number of intersections by drawing lines one by one.

1st line - 0 points=0

2nd line- new 1 point=1

3rd line - old 1 point + new 2 points with two old lines = 1+2

4th line - old (1+2) points+ new 3 points= 1+2+3

5th line - old (1+2+3) points + new 4 points=1+2+3+4

Like that from 35th line-1+2+3+4......34 = \frac{34X35}{2}=595

<u>2nd method:-</u>

We have to choose 2 lines from 35 lines.

We can do it in 35_{C} _{2} ways = 595 points

5 0
4 years ago
Tonya has a wastebasket shaped as a cylinder. The wastebasket measures 20 inches tall and has a radius of 5 inches. 
love history [14]
The amount of space left with the snakd container inside the wastebasket=
=(volume of the wastebasket) - (volume of the sanak container)

Volume (cylinder)=area of the circle * height

1) We calculate the volume of a wastebasket:
area of the circle=πr²=π(5 in)²=25π in²
heigth=20 in.

Volume=25π in² * 20 in=500π in³.

2) we calculate the volume of the snack container:
area of the circle=πr²=π(4 in)²=16π in²
heigth=6 in.

Volume=16π in² * 6 in=96π in³

3) we calculate the amount of space left with the sanak container inside the wastebasket.

amount of space left=500π in³ - 96π in³=404π in³

404π in³=404 * 3.141592654...in³≈1269.2 in³

Answer: 404π in³    or  ≈1269.2 in³
7 0
3 years ago
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