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Brilliant_brown [7]

3 years ago

8

alex sells cars at keith palmer ford. he earns $400 a week plus $150 percar he sells. if he earned $150 last week , how many car

s did he sell?

2 answers:

viktelen [127]3 years ago

5 0

He gets $400 a week plus 150 for each car so he sold only 1 car

Igoryamba3 years ago

3 0

Alex would have sold one car

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Jenny bought gas for her car gas cost 3.45 pee gallon jenny bought 12.2 gallons what was the total cost for jennys gas

Semenov [28]

Answer:

$42.09

Step-by-step explanation:

1. multiply the cost per gallon (3.45) by the gallons purchased (12.2)

3.45*12.2 = 42.09

6 0

3 years ago

Write 27/40 as a decimal.

Alecsey [184]

Answer:

0.675

Step-by-step explanation:

27/40

- Divide each number by 10

2.7/4

- Multiply each number by 25

67.5/100

Convert that into a decimal. :)

4 0

3 years ago

Read 2 more answers

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

5. Patricia is on the edge of the top of a mountain and notices that a kayak on the lake is 30 meters

zhuklara [117]

The distance from the kayak to the mountain to the nearest whole number is 43 meters.

The situation will form a right angle triangle as shown below in the diagram.

<h3>Properties of a right angle triangle:</h3>

One of its angle is 90 degrees.
The sides and angle can be found by using trigonometric ratios.

The height(opposite side) of the triangle is the kayak distance from the mountain to the lake below. The distance from the kayak to the mountain is the adjacent side of the triangle.

Therefore, using trigonometric ratios,

tan 35 = opposite / adjacent

tan 35° = 30 / a

a = 30 / tan 35°

a = 30 / 0.70020753821

a = 42.8444402023

a = 42.84 meters

a ≈ 43 meters

learn more on the angle of depression: brainly.com/question/13969247?referrer=searchResults

8 0

3 years ago

Help please im trying to pass

Stells [14]

Answer:

(0, -1)

Step-by-step explanation:

that is where the line intercepts on the vertical axis

3 0

3 years ago

Read 2 more answers