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2 years ago

Which function below has the lowest y intercept?

Mathematics

1 answer:

amm18122 years ago

3 0

Answer:

g(x)

Step-by-step explanation:

i just took this test and they put the wrong answer and it is actually g(x)

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What is the circumference of the circle shown below?

Wittaler [7]

I think the answer is D but in not sure

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2 years ago

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The volume of a cylindrical container is 226 cubic centimeters. The radius of the container is 6 centimeters. Find the height of

coldgirl [10]

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2 years ago

Joseph owns a pizza restaurant. He sells pizzas for $15 each and charges $8 for delivery. What would your total cost be if you h

ANTONII [103]

Answer:

y = 15x + 8

Step-by-step explanation:

Hey there!

15x represents 15 times how many ever pizzas you order and 8 is for the delivery cost.

Hope this helps!

8 0

2 years ago

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

1 year ago

In triangle △ABC, ∠ABC=90°, BH is an altitude. Find the missing lengths. AB=4 and BC=3, Find AH, CH and BH.

IRISSAK [1]

Answer:

AH = 3.2
CH = 1.8
BH = 2.4

Step-by-step explanation:

It can be convenient to compute the length of the hypotenuse of this triangle (AC). The Pythagorean theorem tells you ...

AC^2 = AB^2 + CB^2

AC^2 = 4^2 + 3^2 = 16 + 9 = 25

AC = √25 = 5

The altitude divides ∆ABC into similar triangles ∆AHB and ∆BHC. The scale factor for ∆AHB is ...

scale factor ∆ABC to ∆AHB = AB/AC = 4/5 = 0.8

And the scale factor to ∆BHC is ...

scale factor ∆ABC to ∆BHC = BC/AC = 3/5 = 0.6

Then the side AH is 0.8·AB = 0.8·4 = 3.2

And the side CH is 0.6·BC = 0.6·3 = 1.8

These two side lengths should add to the length AC = 5, and they do.

The remaining side BH can be found from either scale factor:

BH = AB·0.6 = BC·0.8 = 4·0.6 = 3·0.8 = 2.4

_____

The sides of interest are ...

AH = 3.2

CH = 1.8

BH = 2.4

3 0

2 years ago