What is the middle term of the product of (x + 5)(x + 2)? A.7x B.10x C.18x

1 answer:

8 0

When you use F.O.I.L. on problems you get an answer that looks like this; ax^2 + bx + c. The middle term is bx. To find this term we solve out by multiplication. so x*x + 5*x + 2*x + 5*2 or x^2 + 7x + 10, so the middle term is 7x.

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In which of the following data sets is the mean equal to the mode?

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What is the eighth term of the sequence below?

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Answer:

A. 33

Step-by-step explanation:

Add 4 to each number

1. 5 is your first number

2. 5+4= 9 is your second number

3. 9+4= 13 is your third number

4. 13+4= 17 ...

5. 17+4= 21 ...

6. 21+4= 25 ...

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Which are the solutions of x^2 = –11x + 4?

strojnjashka [21]

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

$x^2 + 11x - 4 = 0$

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}$

where $x_{1,2}$ is a compact way to indicate both solutions $x_1$ and $x_2$ , while $a,b,c$ are the coefficients of the quadratic equation, i.e. we consider the polynomial $ax^2+bx+c$ .

So, in your case, we have $a=1,\ \ b=11,\ \ c=-4$

Plug those values into the formula to get

$x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}$

So, the two solutions are

$x_1 = \frac{-11+\sqrt{137}}{2}$

$x_2 = \frac{-11-\sqrt{137}}{2}$