What is (-2^2)^3 This would be very helpful for finishing this assignment.

What are the center and radius of the circle whose equation is x2 + y2 + 4x = 5?

prisoha [69]

The center is (-2,0)
The radius is -3

5 0

3 years ago

ILL BRAINLIEST YOU PLEASE HELP ME

Margaret [11]

Answer: x = 34

Step-by-step explanation: ok so when you put 2 triangles together and make a rectangle the length and width will still be the same! so the answer is 34ft

8 0

3 years ago

Write an equation that passes through a pair of points

Phoenix [80]

Answer:

C. $y = -x + 2$

Step-by-step explanation:

First, find the slope, m, of the line that passes through the two given points.

Slope = $\frac{rise}{run} = \frac{-2}{2} = -\frac{1}{1} = -1$

Next, determine the y-intercept, b.

The y-intercept is where the line intercepts the y-axis. The line intercepts the y-axis at y = 2. Therefore, b = 2.

Now, to get an equation of the line, substitute m = -1, and b = 2 in $y = mx + b$ .

✅The equation would be:

$y = -1x + 2$

$y = -x + 2$

3 0

3 years ago

How do you solve these equations? I don't want you to answer all of them, just tell me how to solve each type of equation on the

miss Akunina [59]

Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

$\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}$

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$\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}$

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$\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}$