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2 years ago

1.

Mathematics

2 answers:

Degger [83]2 years ago

7 0

Answer: The answer is letter B or letter D

D. quadratic function; quadratic term: −18x2; linear term: −21x; constant term: −6

Step-by-step explanation: Find the degree of the function to determine if linear. f ( x ) = − 18 x 2 − 21 x − 6 is not a linear function

Expand the polynomial using the FOIL method. f ( x ) = − 18 x 2 − 21 x − 6

MrMuchimi2 years ago

3 0

Answer:

Step-by-step explanation:

Got it right on test

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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

The heights of women aged 30 to 35 are approximately Normal with mean 63 inches and standard deviation 2.4 inches. Men the same

gavmur [86]

Answer

given,

mean of women (μ)=63 inches

standard deviation of women (σ)= 2.4 inches

mean of men (μ)=70 inches

standard deviation of women (σ)= 2.5 inches

for women

$P(x>66)$

$P(x>66) = 1-P( \dfrac{x - \mu}{\sigma}$

$P(x>66) = 1-P( z$

using z-table

$P(x>66)= 1 - 0.8944$

$P(x>66)=0.1056[/tex]for men[tex]P(x>66)$

$P(x>66) = 1-P( \dfrac{x - \mu}{\sigma}$

$P(x>66) = 1-P( z$

using z-table

$P(x>66)= 1 - 0.0548$

$P(x>66)=0.9452$

4 0

3 years ago

I will mark as the brainliest

antoniya [11.8K]

A and D because the text states the moon rises from the east and sets in the west, but it can rise earlier in the day when the sun is brighter than the moon so you can't see the moon.

7 0

3 years ago

Read 2 more answers

Help please I don't understand this world problem

Amanda [17]

Multiply each cost by numbers starting from 0 then keep going until you get the same cost for each

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2 years ago

Whats 9+`10+10+10-10

Mkey [24]

Answer:

The answer to this equation is 29 since my other answer got deleted:(