All categories

2 years ago

|2x + 1| - 4 < 5 solve and graph

Mathematics

2 answers:

NeX [460]2 years ago

7 0

Answer:

The value of x could be 3

Step-by-step explanation:

2 x 3 = 6 + 1 - 4 = 3 which is less than 5

Illusion [34]2 years ago

4 0

Answer:

Step-by-step explanation:

Can you give brainliest please

You might be interested in

Scenario: A new Swimming pool is opening up this summer in the Bronx. Students that are 13 - 18 years old can use the pool, gym

MA_775_DIABLO [31]

The answer to your question is 15(12)+40=?
15 being the per month fee, 12 being all 12 months, and 40 being the sign up fee. You can also put 15x12+40=?

3 0

3 years ago

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

3 0

3 years ago

Pls help me ASAP I have other homework to do and I don’t have time for this. It also detects if it’s right or wrong:(

Tems11 [23]

The answer is B. 5 hours.

7 0

2 years ago

Read 2 more answers

Find the measure of Angle A 65°

REY [17]

Answer:

The supplement of 65° is the angle that when added to 65° forms a straight angle (180°).

Step-by-step explanation:

Hope I Helped! :D

7 0

3 years ago

3×

In-s [12.5K]

Answer:

1. x = 39.67

2. x = 15

3. x = 49.29

4. x = -12.8

5. x = 96

6. x = 42

7. x = 36

8. x = 0

9. x = 78

Step-by-step explanation:

Just remember to always isolate the unknown. Here are the solutions to your problem. I will explain each step for the first for you to give you an idea how the others were worked out.

1. $\dfrac{3x}{7}-2 = 15$

Add 2 to both sides to get rid of -2 on the left side.

$\dfrac{3x}{7}-2+(2)=15+(2)///dfrac{3x}{7}=17$

Multiply both sides by 7 to get rid of 7 on the left side.

$\dfrac{3x}{7}\times 7 = 17\times 7\\\\3x = 119$

Divide both sides by 3 to get rid of 3 on the left side.

$\dfrac{3x}{3} = \dfrac{119}{3}\\\\x = 39.67$

You could also transpose everything by the x to the other side of the equation. Just remember that whatever OPERATION used on the original side, must be opposite on the other side. I'll use the second problem to show this.

$\dfrac{2x}{5}+1=7$

Transpose 1 on the left to the right. It is addition on the left, then it would be subtraction on the other side.

$\dfrac{2x}{5}+1=7\\\\\dfrac{2x}{5}=7-1\\\\\dfrac{2x}{5}=6$

Transpose 5 from the left side to the right. It is division on the left, then it would be multiplication on the right.

$\dfrac{2x}{5}=6\\\\2x=6\times 5\\\\2x=30$

Transpose 2 from the left side to the right. It is multiplication on the left, then it would be division on the right.

$2x=30\\\\x=\dfrac{30}{2}\\\\x=15$

Let's move on with the rest now.

$\dfrac{7x}{15}-1=22\\\\\dfrac{7x}{15}=22+1\\\\\dfrac{7x}{15}=23\\\\7x=23\times15\\\\7x=345\\\\x=\dfrac{345}{7}\\\\x=49.29$

$\dfrac{5x}{8}+10=2\\\\\dfrac{5x}{8}=2-10\\\\\dfrac{5x}{8}=-8\\\\5x=-8\times8\\\\5x=-64\\\\x=\dfrac{-64}{5}\\\\x=-12.8$

$\dfrac{x}{6}+4=20\\\\\dfrac{x}{6}=20-4\\\\\dfrac{x}{6}=16\\\\x=16\times6\\\\x=96$

$\dfrac{x}{3}-4=10\\\\\dfrac{x}{3}=10+4\\\\\dfrac{x}{3}=14\\\\x=3\times14\\\\x=42$

$\dfrac{x}{6}+2=8\\\\\dfrac{x}{6}=8-2\\\\\dfrac{x}{6}=6\\\\x=6\times6\\\\x=36$

$\dfrac{x}{9}+8=8\\\\\dfrac{x}{9}=8-8\\\\\dfrac{x}{9}=0\\\\x=0\times9\\\\x=0$

$\dfrac{x}{6}+7=20\\\\\dfrac{x}{6}=20-7\\\\\dfrac{x}{6}=13\\\\x=13\times6\\\\x=78$

6 0

3 years ago