Slope of 10,-1and -8,6

(2-4i)+(3+7i)<br><img src="https://tex.z-dn.net/?f=%282%20-%204i%29%20%2B%20%283%20%2B%207i%29" id="TexFormula1" title="(2 - 4i)

zysi [14]

$(2-4i)+(3+7i)=5+3i$

5 0

4 years ago

Factor the polynomial by it’s greatest common monomial factor. <br> 9b^8 + 24b^3

docker41 [41]

Answer:

9b^8 + 24b^3 = (3b^3) [ 3b^5 + (8)]

greatest common monomial is 3* $b^{3}$

Step-by-step explanation:

we see 9b^8 + 24b^3

there is a common factor here.

3b^3 so

9b^8 + 24b^3 = (3b^3) * 3b^5 + (3b^3) *(8)

9b^8 + 24b^3 = (3b^3) [ 3b^5 + (8)]

so...

8 0

3 years ago

Read 2 more answers

Davina uses a diagram to demonstrate the Pythagorean Theorem.

ElenaW [278]

Answer:

the first option : The area of each square is equal to the square of the length of the side to which it is adjacent.

7 0

3 years ago

A new car is purchased for $17,000 and over time its value depreciates by one half every 3 years. What is the value of the car 1

Jet001 [13]

Answer:

The answer is "$238".

Step-by-step explanation:

Current worth $= \$ 17,000$

depreciates by $\frac{1}{2}$ in 3 years.

time= 19 years

depreciates rate=?

Using formula:

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$\to A_t=A_0(1-\frac{r}{100})^t$

calculates depreciate value in 3 year $= \frac{1}{2} \times 17,000$

$= 8,500$

so,

$A_t=8,500\\\\A_0=17,000\\\\t=3\ years$

$\to A_t=A_0(1-\frac{r}{100})^t\\\\\to 8,500= 17,000(1-\frac{r}{100})^3\\\\\to \frac{8,500}{17,000}= (1-\frac{r}{100})^3\\\\\to \frac{1}{2}= (1-\frac{r}{100})^3\\\\\to (\frac{1}{2})^{\frac{1}{3}}= (1-\frac{r}{100})\\\\\to 0.793700526 = (1-\frac{r}{100})\\\\\to \frac{r}{100} = (1-0.793700526)\\\\\to \frac{r}{100} = (1-0.8)\\\\\to r= 0.2 \times 100 \\\\\to r= 20 \%$

depreciates rate= 20%

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$= \$ 17,000 (1- \frac{20}{100})^{19}\\\\= \$ 17,000 (1-0.2)^{19}\\\\= \$ 17,000 (0.8)^{19}\\\\= \$ 17,000 \times 0.014\\\\= \$ 238$

4 0

3 years ago

Read 2 more answers

Find the slope of each line.

Amiraneli [1.4K]

to get the slope of a line all we need is two points off of it, so let's get two points from that table

$\bf ~\hspace{4em}\downarrow ~\hspace{1.5em}\downarrow \\ \begin{array}{|ccccc|} \cline{1-5} x&0&1&2&3\\ \cline{1-5} y&-\frac{1}{3}&\frac{1}{3}&1&\frac{5}{3} \\ \cline{1-5} \end{array}\qquad \qquad (\stackrel{x_1}{1}~,~\stackrel{y_1}{\frac{1}{3}})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{\frac{5}{3}})$

$\bf \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{\frac{5}{3}}-\stackrel{y1}{\frac{1}{3}}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}}\implies \cfrac{~~\frac{4}{3}~~}{2}\implies \cfrac{~~\frac{4}{3}~~}{\frac{2}{1}}\implies \cfrac{\stackrel{2}{~~\begin{matrix} 4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{3}\cdot \cfrac{1}{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies \cfrac{2}{3}$

6 0

3 years ago