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3 years ago

Find the product. (7−x)(x−1)=

Mathematics

2 answers:

Alexxx [7]3 years ago

5 0

Answer:

-x^2 + 8x - 7

Step-by-step explanation:

(7 - x)(x - 1)

(7 * x) + (7 * -1) + (-x * x) + (-x * -1)

7x - 7 - x^2 + x

8x - 7 - x^2

-x^2 + 8x - 7

Best of Luck!

vaieri [72.5K]3 years ago

3 0

Answer:

-x² + 8x - 7

Step-by-step explanation:

Step 1: Write expression

(7 - x)(x - 1)

Step 2: FOIL (First Outside Inside Last)

7x - 7 - x² + x

Step 3: Combine like terms

-x² + 8x - 7

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If f(x)=x squared-2x and g(x) = 6x+4 for which is a value of x does (f+g)(x) =0

ra1l [238]

Answer:

For $x = -2$

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

In this question:

$f(x) = x^2 - 2x$

$g(x) = 6x + 4$

$f(x) + g(x) = (f+g)(x) = x^2 - 2x + 6x + 4 = x^2 + 4x + 4$

Which is a value of x does (f+g)(x) =0

$x^2 + 4x + 4$

Quadratic equation with $a = 1, b = 4, c = 4$

$\Delta = 4^{2} - 4*1*4 = 0$

$x_{1} = \frac{-4 + \sqrt{0}}{2} = -2$

$x_{2} = \frac{-4 - \sqrt{0}}{2} = -2$

For $x = -2$

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3 years ago

What would be the second part?

Karo-lina-s [1.5K]

Wait what’s the blue part? i might be able to help you with this :)

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3 years ago