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3 years ago

Need to find the greatest common factor

Mathematics

1 answer:

qaws [65]3 years ago

4 0

Answer: The greatest common factor, or GCF, is the greatest factor that divides two numbers. To find the GCF of two numbers: List the prime factors of each number. Multiply those factors both numbers have in common.

Step-by-step explanation:

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Please HELP! I DON'T KNOW HOW CAN I SOLVE THIS

Wewaii [24]

Answer:

Step-by-step explanation:

The Pythagorean Theorem:

$leg^2+leg^2=hypotenuse^2\to a^2+b^2=c^2$

$41.\\\\a=1,\ b=1,\ c=x\\\\x^2=1^2+1^2\\\\x^2=1+1\\\\x^2=2\to x=\sqrt2$

$42.\\a=1, b=2,\ c=x\\\\x^2=1^2+2^2\\\\x^2=1+4\\\\x^2=5\to x=\sqrt5$

$43.\\a=3,\ b=x,\ c=5\\\\3^2+x^2=5^2\\\\9+x^2=25\qquad\text{subtract 9 from both sides}\\\\x^2=16\to x=\sqrt{16}\to x=4$

$44.\\\\a=4,\ b=x,\ c=8\\\\4^2+x^2=8^2\\\\16+x^2=64\qquad\text{subtract 16 from both sides}\\\\x^2=48\to x=\sqrt{48}\\\\x=\sqrt{16\cdot3}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\x=\sqrt{16}\cdot\sqrt3\\\\x=4\sqrt3$

7 0

3 years ago

Read 2 more answers

HELPPP PLSSS

Anastaziya [24]

First place: Liz
Faster: Amanda
Longest time: Steve

4 0

3 years ago

Read 2 more answers

Simplify the rational expression. State any restrictions on the variable.

ExtremeBDS [4]

we are given

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}$

Firstly, we will factor numerators and denominators

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-6)(n^2-4)}{(n^2-3)(n^2-6)}$

we can see that

n^2 -6 is factor on both numerator and denominator

so, it will get cancelled

and n^2 -6 can not be equal to 0

so, one of restriction is

$n^2-6\neq 0$

$n\neq -+ \sqrt{6}$

we can simplify it

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-4)}{(n^2-3)}$

we know that denominator can not be zero

$n^2-3\neq 0$

$n\neq -+ \sqrt{3}$

so, option-B.......Answer

3 0

3 years ago

100 POINTS!!!!!! HELP

irga5000 [103]

A - 1766! I hope this helps

7 0

3 years ago

Please help me I’m desperate

Dafna1 [17]

Answer:

<h2> The first</h2>

Step-by-step explanation:

$\frac29x+3>4\frac59\\{}\quad-3\quad-3\\\\\frac29x\ >\ 1\frac59\\\div\frac29\quad \, \div\frac29\\\\x>1\frac59\div\frac29\\\\x>\frac{14}9\times\frac92\\\\x>\frac{7}1\times\frac11\\\\x>7$

That means every x wich is greater than 7

4 0

3 years ago