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Viktor [21]
3 years ago
6

Whats the answer to m=18+12​

Mathematics
2 answers:
IrinaVladis [17]3 years ago
5 0

Answer:

m = 30

Step-by-step explanation:

Wewaii [24]3 years ago
3 0

Answer:

30

Step-by-step explanation:

10 + 10 = 20

8 + 2 = 10

20 + 10 = 30

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Solve 5kx + 6 = 7kx for x.<br> А. x=3/k<br> в. х = -2/k<br> с. x =-3/k<br> D. x=2/k
slamgirl [31]

Answer:

A.  x = 3/k.

Step-by-step explanation:

5kx + 6 = 7kx

5kx - 5kx + 6 = 7kx - 5kx

6 = 2kx

x = 6 / 2k

x = 3/k.

7 0
2 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}

(c) Concentration of chlorine in the pool at time t

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
Brian ate 2/3 of a pizza. His friend, Joseph, ate 1/4 of what was left. How much pizza did Joseph eat?
Natasha_Volkova [10]

Answer:

1/12

Steps:

1 - 2/3 = 1/3

1/3 / 4 = 1/3 × 1/4 = 1/12

3 0
3 years ago
Reduced form of 2ab^2-a^2 b^2/5
Ket [755]

Answer:

2ab^2-\frac{a^2b^2}{5}=\frac{10ab^2-a^2b^2}{5}

Step-by-step explanation:

Given the expression

2ab^2-\frac{a^2b^2}{5}

\mathrm{Convert\:element\:to\:fraction}:\quad \:2ab^2=\frac{2ab^25}{5}

2ab^2-\frac{a^2b^2}{5}=\frac{2ab^2\cdot \:5}{5}-\frac{a^2b^2}{5}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

                 =\frac{2ab^2\cdot \:5-a^2b^2}{5}

                 =\frac{10ab^2-a^2b^2}{5}

Thus,

2ab^2-\frac{a^2b^2}{5}=\frac{10ab^2-a^2b^2}{5}

7 0
2 years ago
HELLO I REALLY NEED HELP WITH THIS QUESTION:
Ostrovityanka [42]

Answer:

It is the 3rd choice, They didn't multiply by one half

5 0
2 years ago
Read 2 more answers
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