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enyata [817]
3 years ago
11

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:

<h2>41. x = √2</h2><h2>42. x = √5</h2><h2>43. x = 4</h2><h2>44. x = 4√3</h2>

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
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Anastaziya [24]
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4 0
3 years ago
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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
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