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MA_775_DIABLO [31]
3 years ago
13

Help me please. who ever answers gets brainliest​

Mathematics
1 answer:
mezya [45]3 years ago
8 0

Answer:

32

Step-by-step explanation:

(4x6)/2+(2x2)/2+2x2+(2x6)/2

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Anna pays $120 per month for plano lessons. Estimate how much Anna pays
Ivanshal [37]

Answer:

1440

Step-by-step explanation:

1440 because u have multiply 120x12 because there r 12 months

3 0
3 years ago
1) check if they given ordered pairs are solutions to the equation or not.
ziro4ka [17]
Answer: a) Yes
b) No


Explanation: Substitute the values of X and Y from the given points into the equation. I’ll use a) as an example:
Given point: (1,-3)
Equation: 4x-2y=10
Value of X: 1
Value of Y: -3
Substitute X and Y: 4(1)-2(-3)=10
4-(-6)=10
4+6=10 ✔︎

Therefore, the given ordered pair is a proper solution to the equation 4x-2y=10.

*You can use the same method for question b).*
3 0
3 years ago
Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite
zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt  be a gram/L

Let the amount salt in the tank at any time t be Q(t).

\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}

Incoming rate = (a g/L)×(1 L/min)

                       =a g/min

The concentration of salt in the tank at any time t is = \frac{Q(t)}{10}  g/L

Outgoing rate = (\frac{Q(t)}{10} g/L)(1 L/ min) \frac{Q(t)}{10} g/min

\frac{dQ}{dt} = a- \frac{Q(t)}{10}

\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt

Integrating both sides

\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt

\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c        [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c

\Rightarrow -log|10a-15|-2=c

Therefore ,

-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2 .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0  in equation (1) we get

- log|10a|= -log|10a-15| -2

\Rightarrow- log|10a|+log|10a-15|= -2

\Rightarrow log|\frac{10a-15}{10a}|= -2

\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}

\Rightarrow 1-\frac{15}{10a} =e^{-2}

\Rightarrow \frac{15}{10a} =1-e^{-2}

\Rightarrow \frac{3}{2a} =1-e^{-2}

\Rightarrow2a= \frac{3}{1-e^{-2}}

\Rightarrow a = 1.73

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0
3 years ago
Differentiate f(x)=x^2sin(x)
yanalaym [24]

x^2 cos(x)+2x sin (x)

Hope this helps!

6 0
3 years ago
What is the answer to part B? explanation please
vaieri [72.5K]

Answer:

1

Step-by-step explanation:

10³/10³

10³-³

10⁰=1

Formula a¹÷a¹=a¹-¹=a⁰. (Exponents and Powers)

And

a⁰=1

hope it helps..

4 0
3 years ago
Read 2 more answers
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