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Yakvenalex [24]

2 years ago

8

(3x + 1)º Find the value of xWill give Brainlyist

1 answer:

Triss [41]2 years ago

8 0

Answer:

4

Step-by-step explanation:

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Dominik [7]

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Step-by-step explanation:

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3 years ago

*15 Points* Answer the question and I will give you BRAINLIEST!

grandymaker [24]

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8 0

3 years ago

Solve.<br> y=x - 7<br> 5+2y=7

lesantik [10]

Answer:

$\huge\boxed{x=8;\ y=1\to(8;\ 1)}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}y=x-7&(1)\\5+2y=7&(2)\end{array}\right\\\\\text{Substitute (1) to (2) and solve for x:}\\\\5+2(x-7)=7\qquad\text{use the distributive property}\\\\5+(2)(x)+(2)(-7)=7\\\\5+2x-14=7\\\\2x+(5-14)=7\\\\2x-9=7\qquad\text{add 9 to both sides}\\\\2x-9+9=7+9\\\\2x=16\qquad\text{divide both sides by 2}\\\\\dfrac{2x}{2}=\dfrac{16}{2}\\\\\boxed{x=8}$

$\text{Substitute it to (1):}\\\\y=8-7\\\\\boxed{y=1}$

8 0

3 years ago

Read 2 more answers

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

Complete 6,7 for 5 points.

seraphim [82]

Answer:

idk the answer

Step-by-step explanation:

4 0

3 years ago