How to work and get answer for (x-2) (x-6)

1 answer:

7 0

I'm not sure if your multiply or what so i will tell you the ways You have to combine like terms if you are multiplying and you should get
x^2-8x+12
If you are simplifying you have to Expand the polynomial using the FOIL method and you should also get
x^2 -8x+12
Hope this helped please give brainliest.

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Identify the horizontal asymptote of f(x) =x2+5x-3/4x-1

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since the numerator is x² + 5x - 3, and therefore has a degree of 2, whilst the denominator, 4x¹ - 1, has a degree of 1, therefore, there's no horizontal asymptote.

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The number of people signed up for a bus trip increased from 32 to 45. What is the percent increase? Round to the nearest percen

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Step-by-step explanation:

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If the area of a rectangle is 12cm2 what could its length be and its width?

lawyer [7]

Hello,

We know that th formula for the area of a rectangle is: $A=b*h$
Now, the problem in this excercise is that there is no restriction on the values that b and h can take, so, they can take any value provided that it meets: $b*h=12 cm^{2}$
for example:
$lenght= \frac{12}{3}cm \,\,\,\,\,\,and\,\,\,\,width= 3 cm \\ \\ A=\frac{12}{3}cm*3cm \\ A=12 cm^{2}$

But if they can only take intenger values, what we have to do is to descompose 12, and then we can analyze the possible solutions.

12 | 2
6 | 2
3 | 3
1

12= 2 *2*3

first option:
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second option:
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Algebraic Reasoning please help!

stiv31 [10]

Answer:

$\left[\begin{array}{ccc}7\\4\\2\end{array}\right]$

The answer is a single-column matrix (7,4,2)

Step-by-step explanation:

In such multiplication of matrices, you have to proceed by multiplying each ROW of the first matrix by the COLUMN of the second matrix. So,

$\left[\begin{array}{ccc}3&6&1\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (3 * 2) + (6 * 0) + (1 * 1) = 6 + 0 + 1 = 7$

then...

$\left[\begin{array}{ccc}2&4&0\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (2 * 2) + (4 * 0) + (0 * 1) = 4 + 0 + 0 = 4$

and

$\left[\begin{array}{ccc}0&6&2\end{array}\right] * \left[\begin{array}{ccc}2\\0\\1\end{array}\right] = (0 * 2) + (6 * 0) + (2 * 1) = 0 + 0 + 2= 2$

I hope it helps.

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