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statuscvo [17]
2 years ago
10

How many solutions are there?

Mathematics
2 answers:
Oliga [24]2 years ago
8 0

Answer:

no solution

Step-by-step explanation:

Solve:

6(2k + 1) = 12k + 1

Expand brackets:

⇒  12k + 6 = 12k + 1

Subtract 12k from both sides:

⇒  6 = 1

As 6 ≠ 1 there is no solution

maw [93]2 years ago
7 0

Answer:

No solution

Step-by-step explanation:

<u>Simplify the L.H.S and the R.H.S</u>

⇒ 6(2k + 1) = 12k + 1

⇒ 12k + 6 = 12k + 1

<u>Cancel out the "12k"</u>

⇒ 12k + 6 = 12k + 1

⇒ 6 = 1 (\text{False})

Henceforth, this equation has no solution.

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Subtract -13xy from -5xy so<br> -5xy -(-13xy)=?
S_A_V [24]

Answer:

8xy

Step-by-step explanation:

Subtract -13xy from -5xy

-5xy - (-13xy) =

-5xy + 13xy =

8xy

<em>good luck, i hope this helps:)</em>

3 0
3 years ago
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In a mixture 60 liters, the ratio of milk and water 2:1.If this ratio is to be 1:2, then what is the quantity of water to be fur
ella [17]

Answer:

Hello,

60L

Step-by-step explanation:

Quantity of water /Quantity of milk =2/1

Quantity of mixture= Quantity of water +Quantity of milk =60L

Quantity of milk =40 L

Quantity of water =20 L

Let say x the quantity of water to be added

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Flura [38]

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Step-by-step explanation:

5 0
3 years ago
Find the price-demand equation for a particular brand television when the demand is 20 TVs per week at $150 per TV, given that t
Alina [70]

Answer:

~150=50e^{-0.01()20}

~p(100)=\$127.7

Step-by-step explanation:

From the question we are told that:

Price of 20TVs per week P_{20}=\$150

Marginal price-demand function p'(x)=-0.5e-0.01x

Generally the The Marginal price function is mathematically given by

  p'(x)=-0.5e^{-0.01x}  

  p(x)=\int-0.5e^{-0.01x}  

  p(x)=50e^{-0.001x}+C  

Therefore the equation when the demand is 20 TVs per week at $150 per TV

150=50e^{-0.01()20}

Giving

p(x)=50e^{-0.01x}+150-50e^{-0.01(20)}

Therefore the Price when the demand is 100 TVs per week

p(100)=50e^{-0.01(100)}+150-50e^{-0.01(20)}

p(100)=\$127.7

7 0
2 years ago
Help me put in those numbers down below
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