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4 years ago

Identify the vertex and the y-intercept of the graph of the function.

Mathematics

2 answers:

QveST [7]4 years ago

5 0

Answer:

vertex, (4, –2); y-intercept, –6

Step-by-step explanation:

The equation is written in vertex form $y = a(x-h)^2 + k$ where the vertex is (h,k).

The vertex here is (4, -2). This means the solution is vertex, (4, –2); y-intercept, –6.

RoseWind [281]4 years ago

5 0

Answer:

vertex, (4, –2); y-intercept, –6.

Step-by-step explanation:

$y=-0.25(x-4)^{2}-2$

$y=-0.25(x^{2}-8x+16)-2$

$y=-0.25x^{2}+2x-4-2$

$y=-0.25x^{2}+2x-6$

the vertex is $(\frac{-b}{2a},y(\frac{-b}{2a}) )$ where a=-0.25 and b=2. So,

$\frac{-b}{2a}= \frac{-2}{-0.5}=4$

$y(\frac{-b}{2a})=-0.25(4)^{2}+2(4)-6= -4+8-6=-2$

Then, the vertex is (4,-2) and the intercept is y(0)= -0.25(0)+2(0)-6 = -6. So, the answer is the third option.

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4 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

2 years ago