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const2013 [10]
3 years ago
11

What is the radius of 6' circle

Mathematics
1 answer:
Virty [35]3 years ago
8 0
A radius of a 6' circle should be 3.
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Please help me<br> I'm stuck on these questions!
defon
5.) (part 1) is 5x^3, part 2 is 0.25x^2
6.) part 1.) D^2. Part 2.) T^2
7.) part 1.) x^9, part 2.) y^10
9.) part 1.) 15x^2y^3, part 2.) 5ab
8 0
3 years ago
If Heidi is hired by company A, she would be paid $14.60 an hour. If she is hired by company B, she would be paid $16.60 an hour
Alinara [238K]

If Heidi only gets paid when she is talking to clients, she will be working

... 9 × (40 min/60 (min/h)) = 6 h

per day.

Her $2.00 difference in pay per hour will result in a difference in pay per day of

... $2.00/h × 6 h/day = $12/day

Company B will pay Heidi $12 more for a day's work.

5 0
4 years ago
Please answer this correctly
tester [92]
Answer: 1 in the 1-5 range, 4 in the 6-10 range, 1 in the 11-15 range, 0 in the 16-20 range, and 5 in the 21-25 range

Explanation: it’s a histogram, similarity a bar graph, that has its data points lie within a range of values.
4 0
4 years ago
Read 2 more answers
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
4 years ago
5.9 divided by 497.252
romanna [79]

Answer:

0.011865211

Step-by-step explanation:

5.9/497.252 =0.011865211

7 0
3 years ago
Read 2 more answers
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