All categories

2 years ago

Find the value of x

Mathematics

1 answer:

a_sh-v [17]2 years ago

3 0

4x - 7 = 3x + 9

x = 16

they're equal because they are opposites by the vertex.

<em />

<em>hope it helps :)</em>

You might be interested in

Identify holes in the graph of the function.

snow_lady [41]

Answer: x = 0

Step-by-step explanation:

The hole in the graph (a discontinuity) exist where the function doesn't exist. Because anything divided by zero is undefined, then the function would not exit at 0, thus having a hole/discontinuity.

7 0

3 years ago

HALT! DON'T MOVE A MUSCLE! CLICK HERE AND HELP ME OUT?! BRAINLIEST ANSWER INCLUDED! it costs $10 to place an advertisement in th

SashulF [63]

The ordered pairs are probably (0,10) and then (-1,5).

7 0

3 years ago

What is the ratio of 24 packs 2 for $9

topjm [15]

24 cans = 2 cans
$x $9
2x= 260
x=180
24 cans/ $108= 2 cans/ $9

5 0

3 years ago

The base of a rectangular box has a width of 3 inches and a length of 4 inches. The box is 12 inches inches tall. a) Draw a pict

posledela

it would be 144 inches

8 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

Find the value of x​

Find the value of x