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iragen [17]
4 years ago
9

Solve using any strategy. A school band has a brass section of trumpet, trombone, and tuba players. There are twice as many trom

bones as tubas, and half as many trombones as trumpets. If there are two tubas in the band, what is the total number of players in the brass section
A. 8 players


B. 10 players


C. 12 players


D. 14 players
Mathematics
2 answers:
Roman55 [17]4 years ago
7 0
2 Tubas, 4 Trumpets, 8 Trombones, all together you have 14 Players. The answer is D
never [62]4 years ago
6 0
If there are 2 tubas and there are twice as many trombones as there are tubas then there are 4 trombones. And if there are half as many trombones as there are trumpets, then there are 8 trumpets. Therefore, the answer is (D).
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Complete each statement to describe the relationship between gallons and liters.
maw [93]
Conversion
1 gallon = 3.785 liters

To convert gallons to liters multiply gallons by 3.785

1 liter= 0.264172 gallons

To convert liters to gallons multiply liters by 0.264172

3.78541 liters= 1 gallon
6 0
2 years ago
Can someone please help me?
Fed [463]

Answer:

yes

Step-by-step explanation:

3 0
3 years ago
In our family there is five boys and one girl. What percent of the family is boys?
saw5 [17]

Answer:

83% are boys

Step-by-step explanation:

Im not sure if you have to round but you take the total number of family members (6) and divide it by how many boys (5) which would be 0.8333333 which you can round to either 80 or 83

8 0
3 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
Check the alternative in the options below that correspond to a non-equation of the 2nd degree.
PSYCHO15rus [73]
I think it is e and d and I know it is
7 0
4 years ago
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