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4 years ago

Solve for c.

Mathematics

2 answers:

sveticcg [70]4 years ago

5 0

Answer: the correct option is

(D) $c=\dfrac{d+ab}{a}.$

Step-by-step explanation: We are given to solve the following equation for the value of c :

$a(c-b)=d~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

To solve the given equation for c, we need to take c on one side and all other values on the other side of the equation.

The solution for c of equation (i) is as follows :

$a(c-b)=d\\\\\Rightarrow ac-ab=d\\\\\Rightarrow ac=d+ab\\\\\Rightarrow c=\dfrac{d+ab}{a}.$

Thus, the required value of c is $c=\dfrac{d+ab}{a}.$

Option (D) is CORRECT.

ICE Princess25 [194]4 years ago

3 0

a(c−b)=d
ac - ab = d
ac = d + ab
c = (d + ab) / a

answer is
D. c = (d + ab) / a

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A phone company offers two monthly charge plans. In plan a there is no monthly fee, but the customer pays 8 cents per minute of

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$m>2.40$

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Let m denotes the number of minutes of phone use in a month.

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3 years ago

The quantity demanded x each month of Russo Espresso Makers is 250 when the unit price p is $138. The quantity demanded each mon

Stells [14]

Answer:

(a) $D(q)=\frac{-1}{25} q+148$

(b) $S(q)=\frac{1}{50}q+58$

Step-by-step explanation:

(a) For the demand equation D(q) we have

P1: 138 Q1: 250

P2: 108 Q2: 1000

We can find m, which is the slope of the demand equation,

$m=\frac{p_{2} -p_{1} }{q_{2} -q_{1} }=\frac{108-38}{1000-250} =\frac{-30}{750}=\frac{-1}{25}$

and then we find b, which is the point where the curve intersects the y axis.

We can do it by plugging one point and the slope into the line equation form:

$y=mx+b\\\\D(q)=mq+b\\\\138=\frac{-1}{25}(250) +b\\\\138=-10+b\\\\138+10=b=148$

With b: 148 and m: -1/25 we can write our demand equation D(q)

$D(q)=\frac{-1}{25} q+148$

(b) to find the supply equation S(q) we have

P1: 102 Q1: 2200

P2: 102 Q2: 700

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$m=\frac{p_{2} -p_{1} }{q_{2} -q_{1} }=\frac{72-102}{700-2200} =\frac{-30}{-1500}=\frac{1}{50}$

$y=mx+b\\\\D(q)=mq+b\\\\72=\frac{1}{50}(700) +b\\\\72=14+b\\\\72-14=b=58\\$

And we can write our Supply equation S(q):

$S(q)=\frac{1}{50}q+58$

(c) Now we may find the equilibrium quantity q* and the equilibrium price p* by writing D(q)=S(q), which means the demand equals the supply in equilibrium:

$D(q)=S(q)\\\\\frac{-1}{25}q+148=\frac{1}{50}q+58\\\\$

$148-58=\frac{1q}{50} +\frac{1q}{25} \\\\90= \frac{1q}{50} +\frac{2q}{50}\\\\90=\frac{3q}{50}\\ \\q=1500\\\\$

We plug 1500 as q into any equation, in this case S(q) and we get:

$S(q)=\frac{1}{50}q+58\\\\S(1500)=\frac{1}{50}(1500)+58\\\\S(1500)=30+58\\\\S(1500)=88$

Which is the equilibrium price p*.

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