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3 years ago

14

Describe the solution to the system of equations below. HALP

1 answer:

guajiro [1.7K]3 years ago

6 0

Solve for
y
y in
5
x
−
y
=
8
5x−y=8.
y
=
5
x
−
8
y=5x−8

2 Substitute
y
=
5
x
−
8
y=5x−8 into
2
0
x
−
4
y
=
2
4
20x−4y=24.
3
2
=
2
4
32=24

3 Since
3
2
=
2
4
32=24 is not true, this is an inconsistent system.

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What is the solution to log 5 (x+30)= 3

Free_Kalibri [48]

Answer:

x = 95

Step-by-step explanation:

Using the rule of logarithms

• $log_{b}$ x = n ⇔ x = $b^{n}$

Given

$log_{5}$ (x + 30) = 3, then

x + 30 = 5³ = 125 ( subtract 30 from both sides )

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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

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Answer:

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Step-by-step explanation:

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1 cup of trail mix -> X cups of peanuts

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X = 8/27 cups

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Step-by-step explanation:

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