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3 years ago

What value of c makes x2 − 12x + c a perfect square trinomial?

Mathematics

2 answers:

miv72 [106K]3 years ago

6 0

Answer: The required value of c is 36.

Step-by-step explanation: We are given to find the value of c so that the following expression becomes a perfect square trinomial :

$E=x^2-12x+c~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

To find the value of c, we proceed as follows :

$E\\\\=x^2-12x+c\\\\=(x^2-2\times x\times6+6^2)+c-6^2\\\\=(x-6)^2+c-36.$

So, for E to be a perfect square trinomial, we must have

$c-36=0\\\\\Righatrrow c=36.$

Thus, the required value of c is 36.

Nastasia [14]3 years ago

4 0

Answer:

for value 36

c makes the perfect square

x2 - 12x +36

= x2 - 2*6x +6^2

= (x-6)^2 #

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Answer:

Step-by-step explanation:

In order to make a specific shade of green paint, a painter mixes 1 1/2 quarts of blue paint. Converting 1.5 quarts to cup,

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She also used 1/2 gallon of white paint. Converting 0.5 gallons to cup,

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2 years ago

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2 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