0.5x - 0.1 = -2.9

When you multiply each side of an equation by the same number,
you don't change the equation or its solution. If you want to get rid
of the decimals, you could multiply each side by 10. Then the
equation is

 5x - 1 = -29

You should be able to solve that without help.
But I'm about to take your points, so here's
how to solve that equation:

 5x - 1 = -29

Add 1 to each side: 5x = -28

Divide each side by 5 : x = -28 / 5

Simplify the solution: x = -5 and 4/5 .

Again ... this is the same solution as the original equation
if we hadn't gotten rid of the decimals.

5 0

3 years ago

For number 6, evaluate the definite integral.

maks197457 [2]

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\$

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
u(28)=8+2(28)\implies u(28)=64$

$\bf \\\\
-------------------------------\\\\
\displaystyle \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
\\\\\\
\left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2} \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3$

3 0

3 years ago

Complete the table i’ll give brainliest

elena-s [515]

Answer:

1. 4

2. 11

3. 18

4. 25

Step-by-step explanation:

8 0

3 years ago