Whats 9 plus 10? hint: its not 19

Leona [35]

You would multiply it by 4
4(3x^2 +2 )
12x^2 +8

6 0

3 years ago

The following circle graph represents how the Townsends spend their monthly income. What is the central angle measure of the sec

mars1129 [50]

Please say brainliest.

It would be 18 degrees.

5% of 360(the total number of degrees in a circle) = 18 degrees

3 0

3 years ago

Donna wants to decorate her bulletin board she has 38 meters of green ribbon to go around the bulletin board and she has to use

andreev551 [17]

If you are asking for the perimeter of the bulletin board the answer below will be correct!

Answer:

Square option: 9.5 Meters

Rectangle option: see explanation below

Step-by-step explanation:

To find the perimeter you must add the lengths of all the sides. Assuming the bulletin board is a SQUARE all four sides are going to be equal. Therefore you can divide 38 by 4 to get the length of one side (aka 9.5 meters).

Assuming the bulletin board is a RECTANGLE changes things up. You can pretty much pick any combination of numbers that will add up to 19 (because if it is a rectangle there are two pairs of equal side lengths). You could do 1 + 18, 5+14, 8+11. The options are really endless

4 0

2 years ago

Someone help pls :CCCCC its math

Nutka1998 [239]

6x3=18

The room is 18ft long

100/2=50

The park is 50mm long

3 0

3 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

3 years ago