Plsssss help meeeeeee!!

PLEASE HELP!!!!!! In which week was the rainfall in Portland the highest and how much fell?

garik1379 [7]

Answer:

Week 4, 1400

Brainliest plz

Step-by-step explanation:

5 0

3 years ago

Help on 3. And 4. Thank you!

chubhunter [2.5K]

Answer:

#3.) The initial value of 16 gal at x = 5 minutes means that 16 gallons of water was present 5 minutes after the barrel started leaking.

#4.) Find how many minutes until the barrel is empty of water.

Let y = 0, to solve for time x.

0 = (-2/5)*x + 18

(2/5)*x = 18

x = (5/2)* 18 = 5*9 = 45 minutes

Up to and after 45 minutes, the barrel is empty of water.

Step-by-step explanation:

#2.)

minutes: 5, 10, 15, 20

water(gal): 16, 14, 12, 10

Find slope: slope m = (14 - 16)/(10 - 5) = -2/5

y - 10 = (-2/5)*(x - 20)

y - 10 = (-2/5)* x + 8

y = (-2/5)*x + 18

rate of change slope means that for every minute 2/5 gallons of water is lost

#3.) The initial value of 16 gal at x = 5 minutes means that 16 gallons of water was present 5 minutes after the barrel started leaking.

#4.) Find how many minutes until the barrel is empty of water.

Let y = 0, to solve for time x.

0 = (-2/5)*x + 18

(2/5)*x = 18

x = (5/2)* 18 = 5*9 = 45 minutes

4 0

4 years ago

Solve for H<br><br> 19+4H=1-5H

Jobisdone [24]

19=1-5h-4h
19-1=-5h-4h
18=-9h
18/-9=h
h=-0.5

7 0

3 years ago

Select the pair of slopes that are the slopes of perpendicular lines.

agasfer [191]

<h3> $\frac{3}{4}\ and\ \frac{-4}{3}$ are the slopes of perpendicular lines</h3>

<em><u>Solution:</u></em>

We know that,

Product of slope of a line and slope of line perpendicular to it is always equal to -1

<em><u>Option 1</u></em>

$\frac{1}{2}\ and\ 2$

$Product = \frac{1}{2} \times 2 = 1$

Thus, these are not the slopes of perpendicular lines

<em><u>Option 2</u></em>

$\frac{3}{4}\ and\ \frac{-4}{3}\\\\Product = \frac{3}{4} \times \frac{-4}{3}\\\\Product = -1$

This option satisfies the slopes of perpendicular lines

<em><u>Option 3</u></em>

$\frac{-2}{5}\ and\ \frac{-5}{2}\\\\Product = \frac{-2}{5} \times \frac{-5}{2} = 1$

Thus, these are not the slopes of perpendicular lines

<em><u>Option 4</u></em>

2 and 2

Product = 2 x 2 = 4

Thus, these are not the slopes of perpendicular lines

7 0

3 years ago

Ava and Kelly ran a road race, starting from the same place at the same time. Ava ran at an average speed of 6 miles per hour. K

Zepler [3.9K]

<h2>Hello!</h2>

The answer is:

Ava and Kelly will be 3/4 mile apart after 0.375 hours.

To calculate when will Ava and Kelly be 3/4 mile apart, we need to write the equations for both Ava's and Kelly's positions.

Writing the equations we have:

For Ava:

$x_A=x_o+v_ot\\\\x_A=0+v_o*t\\\\x_A=v_{o(ava)*t$

For Kelly:

We need to calculate when Kelly will be 3/4 mile apart becase is running faster than Ava, so, writing the equation we have:

$x_K=x_A+\frac{3}{4}mile=x_o+v_{o(Kelly)}*t$

Now, substituting the equation for Ava into the equation for Kelly, we have:

$x_A+\frac{3}{4}mile=x_o+v_{o(Kelly)}*t$

$v_{o(ava)*t+\frac{3}{4}mile}=x_o+v_{o(Kelly)}*t$

$v_{o(ava)*t+0.75mile}=x_{o}+v_{o(Kelly)}*t\\\\6mph*t+0.75mile=8mph*t\\\\0.75mile=2mph*t\\\\t=\frac{0.75miles}{2mph}=0.375hours$

To prove that the result is correct, we just need to substitute the obtained value for time into both equations, so, substutiting we have:

For Ava:

$x_A=v_{o(ava)*t=6mph*0.375hours=2.25miles$

For Kelly:

$x_K=v_{o(Kelly)}*t=8mph*0.375=3miles$

There is a difference of 0.75 miles or 3/4 mile between Ava and Kelly, so, the obtained value for time is correct.

Therefore, we have that Ava and Kelly will be 3/4 mile apart after 0.375 hours.

Have a nice day!

5 0

3 years ago