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3 years ago

Please describe to me why there are no middle terms when multiplying (x - 4) and (x + 4)

Mathematics

1 answer:

julsineya [31]3 years ago

7 0

Answer:

The middle terms cancel out

Step-by-step explanation:

FOIL the problem:

(x-4) * (x+4)

There are two middle terms, one positive and one negative:

x^2 + 4x - 4x -16

The positive and negative 4s cancel and you are left with just the first and last term:

x^2 - 16

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Write an algebraic expression for each of the statements below.

4vir4ik [10]

Answer:

i) 3x + 7

ii) x/3 = 9

iii) y = 2x +4

iv) y = 6x

5 0

3 years ago

Write an equation of the line perpendicular to

AysviL [449]

Answer:

From general equation of a line:

$y = mx + b$

m » slope, b » y-intercept.

but perpendicularly:

$m _{1} \times m _{2} = - 1 \\ - 3 \times m = - 1 \\ m = \frac{1}{3}$

at point (6, 2):

$2 = ( \frac{1}{3} \times 6) + b \\ 2 = 2 + b \\ b = 0$

therefore, equation is :

${ \boxed{ \boxed{y = \frac{1}{3} x}}}$

8 0

3 years ago

PLEASE HELP ME PLEASEEEEE

Feliz [49]

Our first step is to get rid of the fraction by multiplying both sides by 3.

So in our second step we have x - 4 > -36.

Now add 4 to both sides to get x > -32.

Since x is greater than -32, the values that would

make this inequality true are -31 and 31.

6 0

3 years ago

BRAINLIEST! Number 4 please!

Ratling [72]

I would say c because they both are the same thing just on different sides/ rotated

6 0

3 years ago

3x^2 – 2x + 7 = 0, then (x -1/3)^2

loris [4]

$3x^{2} - 2x + 7 = 0 \\ \\ y = (x- \frac{1}{3})^{2} \\ \\ y = x^{2} - \frac{2}{3}x + \frac{1}{9} \\ \\ 3y = 3x^{2} - 2x + \frac{1}{3} \\ \\ 3y = 3x^{2} - 2x + 7 - \frac{20}{3} \\ \\ 3y = 0 - \frac{20}{3} \\ \\ 3y = - \frac{20}{3} \\ \\ \boxed{y =-\frac{20}{9}}$

3 0

4 years ago