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4 years ago

I need step by step explanation for thisss >33

Mathematics

2 answers:

Diano4ka-milaya [45]4 years ago

5 0

find the number eight times of a two digit number higher than the number nineteen. The tens digit than the units digit .you have to multiply 8 to a 2 digit number exceeding nineteen so you find a number exceeding ninteen which would have to be able to to times by 8 to go higher than the number nineteen

Debora [2.8K]4 years ago

3 0

Answer:

Step-by-step explanation:

The question says find the number eight times of a certain two digit number exceeding the number 19.The tens digit than the units digit . So,you have to multiply eight to a two digit number exceeding 19 so you find a number exceeding 19 which would have to be able to to multiply by eight to exceed the number 19

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.If x2 + kx+ 6 = (x + 2) (x + 3) for all x, then the value of k is<br><br>

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Step-by-step explanation:

$(x + 2)(x + 3) = {x}^{2} + 5x + 6$

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3 years ago

I WILL GIVE YOU 40 POINTS PLEASE HELP<br> (2)/(x^2-9) - (3x)/( x^2-5x+6) please show work

mars1129 [50]

Answer:

$\frac{-3x^2-7x-4}{x^3-2x^2-9x+18}$

Step-by-step explanation:

$\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factor x²-9 and x²-5x+6.

$\frac{2}{\left(x+3\right)\left(x-3\right)}-\frac{3x}{\left(x-2\right)\left(x-3\right)}$

Least common multiple of (x+3), (x-3), (x-2), and (x-3) is (x+3), (x-3), and (x-2).

Adjust the fractions based on LCM.

$\frac{2\left(x-2\right)}{\left(x+3\right)\left(x-3\right)\left(x-2\right)}-\frac{3x\left(x+3\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}$

Subtract the fractions since denominators are equal.

$\frac{2\left(x-2\right)-3x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x-2\right)}$

Expand.

$\frac{-3x^2-7x-4}{x^3-2x^2-9x+18}$

The fraction can be in factored form.

$\frac{-\left(x+1\right)\left(3x+4\right)}{\left(x-2\right)\left(x+3\right)\left(x-3\right)}$

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4 years ago