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3 years ago

10

There are 21 students in Mr.Tentlers class The students divided themselves evenly into 3 groups.how many students are in ea

ch groups?

2 answers:

iVinArrow [24]3 years ago

6 0

$21:3=7 \ students \ are \ in \ each \ group$

arsen [322]3 years ago

4 0

There are 7 in each grops because if 3x7 is 21 then 21 divide by 3 is 7

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I Need help! If you help me you'll get 10 points!!

natka813 [3]

To solve this problem you must apply the proccedure shown below:

1. You have that the number of innings are expressed as a mixed number in the exercise:

$102^{\frac{2}{3} }$

2. If you want to write it as a decimal, you can convert the fraction as a decimal by dividing the numerator by the denominator and then, you must add this to the whole number part:

$102+\frac{2}{3}=102+0.67=102.67$

The answer is: 102.67

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3 years ago

You were planning to spend $12 on a

nordsb [41]

You have $12, and need at least x more dollars to spend $30 total. So the inequality is 12 + x ≥ 30, where x is the number of dollars needed to get free delivery. To solve for x, subtract 12 from each side of the sign. x ≥ 30 - 12, x ≥ 18. You must spend at least 18 more dollars to qualify for free delivery.

8 0

2 years ago

Two cans of paint and 1 brush cost $37. Four cans of paint and 3 brushes cost $81. Find the cost of one can of paint.

weeeeeb [17]

2c + b = 37 ....multiply by -3
4c + 3b = 81
------------------
-6c - 3b = - 111 (result of multiplying by -3)
4c + 3b = 81
------------------add
-2c = - 30
c = -30/-2
c = 15 <== $ 15 for a can of paint

and it costs $ 7 for each brush

8 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

4 0

3 years ago

Need help again ! answer asap this is so hard

Whitepunk [10]

Answer:

I think that C is the only one that is not Trigonometric.

Step-by-step explanation:

8 0

3 years ago