All categories

3 years ago

-7 - 7 ( 8x - 7 ) = - 19 + 5x

Mathematics

1 answer:

Yakvenalex [24]3 years ago

6 0

-7 - 7( 8x - 7 ) = -19 + 5x
⇔ -56x + 49 = -12 + 5x
⇔ -61x = -61
⇔ x = 1

You might be interested in

How to do these first four please! I believe graphing quadratics thx!! Need ASAP

sergeinik [125]

Graphed it
If you need to do it manualy, consider using a table of values

5 0

3 years ago

Help help help help help help

sammy [17]

Answer:

Explanation:

7. f(x) = x^4 + 3x^3 - 6x^2 - 10x + 8
f(2) = 2^4 + 2^3 - 6^2 - 10(2) + 8
f(2) = 16 + 8 - 12 - 20 + 8
f(2) = 0

8. x^2 - 9 = (x - 3)(x + 3)
x ^2 + x - 6 = (x + 3)(x -2)

Thus, the common factor is x + 3

5 0

2 years ago

heights for 16 year old boys are normally distributed with a mean if 68.3 in and a standard deviation of 2.9 in. find the z-scor

allsm [11]

Simple just decide pie by the universe multiplied bye the square rout of a f1 race car most likely leading too yes

6 0

3 years ago

A cross-country racer travels 20 kilometers before she realizes that she has to cover 75 kilometers in order to qualify for the

Katarina [22]

6............................................

5 0

3 years ago

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