Solve the equation for x.

Marvin solved the equation 4(x+5)=88 fill in the blanks

enot [183]

Answer:

x=17

Step-by-step explanation:

4(x+5) = 88

4x + 20 = 88

4x = 88-20

4x = 68

x = 68/4

x = 17

7 0

3 years ago

Read 2 more answers

Solve the logarithmic equation . Round the answer to four decimal places . 2^x =25

levacccp [35]

Answer:

x = 4.6439

Step-by-step explanation:

2^x =25

Take the log of each side

log 2^x = log 25

We know that log a^b = b log a

x log 2 = log 25

Divide each side by log 2

x = log 25 / log 2

x =4.643856

Rounding to 4 decimal places

x = 4.6439

8 0

3 years ago

Read 2 more answers

Find an expression which represents the difference when (6x-3y) is subtracted from (9x+2y) in simplest terms.

Alex73 [517]

Answer:

$3x+5y$

Step-by-step explanation:

Subtract (6x-3y) from (9x+2y):

$(9x+2y)-(6x-3y)$

This can also be seen as:

$(9x+2y)-1(6x-3y)$

Distribute the -1 to (6x-3y):

$(9x+2y)-1(6x)-1(-3y)\\\\(9x+2y)-6x+3y$

Simplify the parentheses:

$9x+2y-6x+3y$

Group like terms:

$(9x-6x)+(2y+3y)$

Combine:

$3x+5y$

This expression can't be further simplified. Therefore, 3x+5y is the answer.

:Done

6 0

2 years ago

Help with this question please

ahrayia [7]

20/8=2’5”
So the larger triangle is bigger by 2.5
So you times 5’6” by 2.5 to get your answer which is 14 ft

Hope this helps

3 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