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3 years ago

How do you solve this multi-step equation? -4x-7-3x+4=25

Mathematics

1 answer:

kirill115 [55]3 years ago

5 0

Answer:

x= -4

Step-by-step explanation:

Combine alike terms:

-7x-3=25

Add a 3 to both sides:

-7x=28

Divide by -7 on both sides:

x= -4

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Y=-3x + 10<br> y = -2x + 6<br> Substitution method

Nutka1998 [239]

Answer:

x=4 and y=−2

Step-by-step explanation:

y=−3x+10;y=−2x+6

Step: Solve y=−3x+10for y:

Step: Substitute−3x+10foryiny=−2x+6:

y=−2x+6

−3x+10=−2x+6

−3x+10+2x=−2x+6+2x(Add 2x to both sides)

−x+10=6

−x+10+−10=6+−10(Add -10 to both sides)

−x=−4

-x/-1 = -4/-1 (Divide both sides by -1)

x=4

Step: Substitute4forxiny=−3x+10:

y=−3x+10

y=(−3)(4)+10

y=−2(Simplify both sides of the equation)

7 0

2 years ago

Read 2 more answers

How many lines of symetry do a heart have

Deffense [45]

Answer:

Step-by-step explanation:

You can only cut it in half (vertically)

7 0

2 years ago

A teacher is creating boxes of supplies for her classroom. She places 4 markers in each box and leaves 12 markers at her desk to

polet [3.4K]

Answer:

x =10

Step-by-step explanation:

52 - 12 = 40

40 divide 4 = 10

she has 10 boxes

6 0

2 years ago

Pergunta 5 You need to fill a football with air to play with it. You know that your pump expels air at speed of 8.2 ft/s. The ne

ra1l [238]

The volume flow rate of the air being pumped into the football will be given by:
Q=V*A
Where
V=Speed
A=Area
You know that your pump expels air at speed of 8.2 ft/s
The area of The needle of your pump is
A=pi*r^2=pi*((4.5/1000)*(3.28))^2=0.00068442 ft^2
Finally
Q=V*A=(8.2 ft/s)*(0.00068442ft^2)=0.005612243 ft^3/s

8 0

2 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

2 years ago