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3 years ago

24x-12x^2=0 What is the result of this(please step by step)

Mathematics

1 answer:

Sav [38]3 years ago

5 0

Okay..
12 x^2 = 24x

12x = 24

x= 2 and x=0

hope it helped

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What is the length of the line TR

8090 [49]

Answer:

Step-by-step explanation:

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2.5 * 6 is 15

so 15 is the length of TR

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Which graph shows this system of equations?

SashulF [63]

Answer:

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3 years ago

What is the greatest common factor of 44c^5, 22c^3 and 11c^4

Alexxandr [17]

$\text{ GCF of } 44c^5, 22c^3 , 11c^4 \text{ is } 11c^3$

Solution:

Given that, we have to find the greatest common factor of $44c^5, 22c^3 , 11c^4$

The greatest number that is a factor of two (or more) other numbers.

When we find all the factors of two or more numbers, and some factors are the same ("common"), then the largest of those common factors is the Greatest Common Factor

$\underline{\text{ GCF of } 44c^5, 22c^3 , 11c^4 :}$

First find the GCF of numbers and then find the GCF of variables and then multiply them together

Step 1: GCF of 44, 22 and 11

The factors of 11 are: 1, 11

The factors of 22 are: 1, 2, 11, 22

The factors of 44 are: 1, 2, 4, 11, 22, 44

11 is the greatest factor that is common in above three list

Then the greatest common factor is 11

$\text{ Step 2: GCF of }c^5, c^3 , c^4$

$c^5 = c^3 \times c^2\\\\c^3 = c^2 \times c^1 \text{ or } c^3\\\\c^4 = c^3 \times c^1$

Thus the greatest common factor is $c^3$

Step 3: Multiply the GCF of 44, 22, 11 and GCF of $c^5, c^3 , c^4$

$\text{ GCF of } 44c^5, 22c^3 , 11c^4 = 11 \times c^3 = 11c^3$

Thus $\text{ GCF of } 44c^5, 22c^3 , 11c^4 \text{ is } 11c^3$

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3 years ago

Two trains leave stations 288 miles apart at the same time and travel towards each other. One train travels at 95 miles per hour

erma4kov [3.2K]

95x=85x+288
You could add 288 to any side, it just shows that they're 288 miles apart.
Let's subtract 85x from both sides.
10x=288
28.8=x
It'll take 28.8 hours for the trains to meet. Let's make sure.
28.8×95
2,736
28.8×85
2,448
2,736-2,448
288
So, they'll meet up after 28.8 miles.

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jeka57 [31]

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