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1 year ago

Multiply 2 2/3 and 1 5/6 Simplify the answer and write as a mixed number.

Mathematics

1 answer:

fiasKO [112]1 year ago

4 0

The answer as a mixed number is 4 8/9

The first step is to write out the fractions and multiply both fractions

2 2/3 × 1 5/6

= 8/3 × 11/6

= 44/9

Convert to a mixed number

= 4 8/9 (as a mixed number)

Hence the mixed number gotten from the simplification is 4 8/9

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Which equation is equivalent to log Subscript 3 Baseline (x 5) = 2?.

grin007 [14]

The equivalent equation is x - 4 = 0.

<h2>Linear system</h2>

It is a system of an equation in which the highest power of the variable is always 1. A one-dimension figure that has no width. It is a combination of infinite points side by side.

<h3>Simplification</h3>

Simplification is to make something easier to do or understand and to make something less complicated.

Given

The expression is $\rm log_3(x + 5) = 2$ .

The equivalent equation.

<h3>How to find the equivalent equation.</h3>

The expression is $\rm log_3(x + 5) = 2$ .

On simplifying. we get

$\begin{aligned} \rm log_3(x + 5) &= 2\\x + 5 &= 3^2\\x + 5 &= 9 \\x &= 9 - 5\\x &= 4\\x - 4 &= 0\\\end{aligned}$

The equivalent equation is x - 4 = 0.

More about the linear system link is given below.

brainly.com/question/20379472

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2 years ago

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3 years ago

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2 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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3 years ago

How many square corners does a pentagon have

SIZIF [17.4K]

I assume that by square corners you mean 90 degree angles. if this is the case then there are 2 in a pentagon. Hope I helped :)

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3 years ago