Can some one help me plz

What is the slope of the line that passes through the points (-2,-12) and (1,-3)?

Nookie1986 [14]

Answer:

3

Step-by-step explanation:

Change in Y/Change in X = Slope

8 0

4 years ago

Read 2 more answers

What is the solution to the equation One-third (x minus 2) = one-fifth (x + 4) + 2? x = 12 x = 14 x = 16 x = 26

Elena L [17]

Answer:

x = 26

Step-by-step explanation:

Given

$\frac{1}{3}$ (x - 2) = $\frac{1}{5}$ (x + 4) + 2

Multiply through by 15 to clear the fractions

5(x - 2) =3(x + 4) + 30 ← distribute and simplify both sides

5x - 10 = 3x + 12 + 30, that is

5x - 10 = 3x + 42 ( subtract 3x from both sides )

2x - 10 = 42 ( add 10 to both sides )

2x = 52 ( divide both sides by 2 )

x = 26

3 0

3 years ago

Jane spent 3 hours exploring a mountain with a dirt bike. First, she rode 55 miles uphill. After she reached the peak she rode f

Neko [114]

Answer:

Her speed on the summit was 35 mph.

Step-by-step explanation:

Her speed on the summit was "x" mph while her speed while climbing was "x - 10" mph. The distance she rode uphill was 55 miles and on the summit it was 28 miles. The total time she explored the mountain was 3 hours. Therefore:

time uphill = distance uphill / speed uphill = 55 / (x - 10)

time summit = distance summit / speed summit = 28 / x

total time = time uphill + time summit

3 = [55 / (x - 10)] + 28 / x

3 = [55*x + 28*(x - 10)]/[x*(x - 10)]

3*x*(x - 10) = 55*x + 28*x - 280

3x² - 30*x = 83*x - 280

3x² - 113*x + 280 = 0

x1 = {-(-113) + sqrt[(-113)² - 4*(3)*(280)]}/(2*3) = 35 mph

x2 = {-(-113) - sqrt[(-113)² - 4*(3)*(280)]}/(2*3) = 2.67 mph

Since her speed on the uphill couldn't be negative the speed on the summit can only be 35 mph.

6 0

3 years ago

Need help with number 3

marin [14]

IM THINKING B.)
HOPE ITS RIGHT LOL

3 0

3 years ago

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

4 years ago