What multiplication fact could you use to find a number that can be divided evenly by 8 and by 9

Mathematics

2 answers:

Ad libitum [116K]3 years ago

8 0

72 would be a good one

earnstyle [38]3 years ago

8 0

For 8 you can use 64
For 9 you can use 81
Just multiply 8 and 9 by themselves

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For two functions, a(x) and b(x), a statement is made that a(x) = b(x) at x = 2. What is definitely true about x = 2? (2 points)

Irina-Kira [14]

The statement that is definitely true about x = 2 is Both a(x) and b(x) have the same output value at x = 2.

<h3>How to find the statement that is true about the functions a(x) = b(x) at x = 2?</h3>

If we have two functions a(x) and b(x), a statement is made that a(x) = b(x) at x = a, this implies that the values of the functions a(x) and b(x) are equal at x = a.

Given that the two functions a(x) and b(x), a statement is made that a(x) = b(x) at x = 2.

Then the statement that is definitely true about x = 2 is Both a(x) and b(x) have the same output value at x = 2.

So, the statement that is definitely true about x = 2 is Both a(x) and b(x) have the same output value at x = 2.

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8 0

1 year ago

HELP ME QUICK PLEASEEEE solve for x: 2/x-2+7/x^2-4=5/x

Anna35 [415]

$\displaystyle\\\\\frac{2}{x-2}+\frac{7}{x^2-4}=\frac{5}{x}\\\\\frac{^{x+2)}2~~~~~}{x-2}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2(x+2)}{(x-2)(x+2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2x+4+7}{x^2-4}=\frac{5}{x}\\\\\frac{2x+11}{x^2-4}=\frac{5}{x}\\\\5(x^2-4)=x(2x+11)\\\\5x^2-20=2x^2+11x\\\\5x^2-2x^2 -11x-20=0$

$\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}$