I really need help please help me out!

Use the distance to find the distance between the points A (-3,7 and B (-2,5).

Eva8 [605]

Step-by-step explanation:

1 2 3 4 5

wa

6 0

3 years ago

What’s the midpoint P(0,-2), Q(1,1

LiRa [457]

You would have to add x1 + x2 and divide by 2. do the same with y1 and y2.

(1/2,-1/2)

6 0

3 years ago

In the proportion 1/z = 4/5/8 which number os equal to z in the proportion

kobusy [5.1K]

I think it’s -156.250771950772

6 0

3 years ago

Which ordered pairs make both inequalities true? Check all that apply.

Thepotemich [5.8K]

All others did not satisfy the inequality simultaneously except #4 and #6
For #4:
2 < 5(1) + 2 i.e. 2 < 7 (true)
2 >= 1/2(1) + 1 i.e. 2 >= 3/2 (true)
For #6
2 < 5(2) + 2 i.e. 2 < 12 (true)
2 >= 1/2(2) + 1 i.e. 2 >= 2

7 0

3 years ago

Read 2 more answers

Evaluate the following limit:

Makovka662 [10]

If we evaluate the function at infinity, we can immediately see that:

$\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = \lim_{x \to \infty}{\frac{(x^2 + 1)^2 - 3x^2 + 3}{x^3 - 5}} = \frac{\infty}{\infty}} \end{gathered}$}$

Therefore, we must perform an algebraic manipulation in order to get rid of the indeterminacy.

We can solve this limit in two ways.

By comparison of infinities:

We first expand the binomial squared, so we get

$\large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_{x \to \infty}{\frac{x^4 - x^2 + 4}{x^3 - 5}} = \infty \end{gathered}$}$

Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.

Dividing numerator and denominator by the term of highest degree:

$\large\displaystyle\text{$\begin{gathered}\sf L = \lim_{x \to \infty}\frac{x^{4}-x^{2} +4 }{x^{3}-5 } \end{gathered}$}\\$

$\ \ = \lim_{x \to \infty\frac{\frac{x^{4} }{x^{4} }-\frac{x^{2} }{x^{4}}+\frac{4}{x^{4} } }{\frac{x^{3} }{x^{4}}-\frac{5}{x^{4}} } }$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{=\lim_{x \to \infty}\frac{1-\frac{1}{x^{2} } +\frac{4}{x^{4} } }{\frac{1}{x}-\frac{5}{x^{4} } } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{0}=\infty } \end{gathered}$}$

Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.

5 0

2 years ago