Marcus bought a phone on sale that was discounted 1/6 the

Margaret is a waitress at The Seafood Bonanza. She earns $3 an hour plus 90% of her tips. Write an equation for her salary, S, i

frosja888 [35]

Answer: 3h + 0.9s

Step-by-step explanation:

90% is the same as 0.9 of..

6 0

3 years ago

Find the value of r so the line that passes through (-5,2) and (3,r) has a slope of -1/2

soldier1979 [14.2K]

The value of r so the line that passes through (-5,2) and (3,r) has a slope of -1/2 is -2

Solution:

Given that line is passing through point (-5, 2) and (3, r)

Slope of the line is $\frac{-1}{2}$

Need to determine value of r.

Slope of a line passing through point $\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right)$ is given by following formula:

$\text { Slope } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ --- eqn 1

$\text { In our case } x_{1}=-5, y_{1}=2, x_{2}=3, y_{2}=\mathrm{r} \text { and } m=-\frac{1}{2}$

On substituting the given value in (1) we get

$\begin{array}{l}{-\frac{1}{2}=\frac{r-2}{3-(-5)}} \\\\ {\text { Solving the above expression to get value of } r} \\\\ {=>-\frac{1}{2}=\frac{r-2}{3+5}} \\\\ {=>-8=\frac{r-2}{3+5}} \\\\ {=>-8=2(r-2)} \\\\ {=>-8=2 r-4} \\\\ {=>2 r=-8+4} \\\\ {=>2 r=-4} \\\\ {=>r=\frac{-4}{2}=-2}\end{array}$

Hence the value of "r" is -2

8 0

3 years ago

the factor tree for 60 has all of the factors in the factor trees for 15 and 30, but not 45. explain why this is true

adoni [48]

15 and 30 are factors of 60 but 45 is not

4 0

3 years ago

Read 2 more answers

Which is the approximate solution to the system y = 0.5x + 3.5 and y = − 2/3 x + 1/3 shown on the graph? (–2.7, 2.1) (–2.1, 2.7)

AlekseyPX

Answer:

The approximate solution to the system is $(-2.7, 2.1)$ .

Step-by-step explanation:

To solve the system of equations $\begin{bmatrix}y=0.5x+3.5\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$ you must:

$\mathrm{Rationalize\:equations}\\\\\begin{bmatrix}y=\left(\frac{1}{2}\right)x+\left(\frac{7}{2}\right)\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$

$\mathrm{Subsititute\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\begin{bmatrix}-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\\\\-\frac{2}{3}x=\frac{19}{6}+\frac{1}{2}x\\\\-\frac{7}{6}x=\frac{19}{6}\\\\6\left(-\frac{7}{6}x\right)=\frac{19\cdot \:6}{6}\\\\-7x=19\\\\x=-\frac{19}{7}\approx-2.7$

$\mathrm{For\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\mathrm{Subsititute\:}x=-\frac{19}{7}\\\\y=-\frac{2}{3}\left(-\frac{19}{7}\right)+\frac{1}{3}\\\\y=\frac{15}{7}\approx 2.1$