What is the radius of 6' circle

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}}$