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Vilka [71]
3 years ago
10

If you know a point on a line and you know the equation of a line parallel to this line, explain how to write the line’s equatio

n.
Mathematics
1 answer:
vesna_86 [32]3 years ago
7 0

Answer:

Step-by-step explanation:

First, parallel lines have the same slope, so the slopes should be the same , and then you use the given point to find the y intercept

EX:

y=x+1

find equation of line the is parallel and passes through (1, 1)

Same slope, so y=x+b

then plug in the point

1=1+b

b=0

our line would be y=x

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Sketch the graph for each function. Choose either A, B, C, or D.
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Hi. Here it is!
I hope this help...

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3 years ago
8th grade math, please help
Marrrta [24]

I can't tell what it wants to say in the first part, but I'll do A and B for you.

Because all triangle angles add to 180 degrees, you know that angle PMU is 180-21+35 which is 124.

So a) is 124 degrees

And for b) you take the straight line and you know its 180 degrees. Knowing that angle UMP adds up to 180 degrees when added to PMU which is 124, then you subtract 180 - 124

So b) is 56 degrees.

Hope this helps

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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

(c)p_{*} =88\\\\q_{*} =1500

Step-by-step explanation:

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

<em>P1: 138 Q1: 250</em>

<em>P2: 108 Q2: 1000</em>

We can find <u><em>m</em></u>, 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

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

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

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

<em>P1: 102 Q1: 2200</em>

<em>P2: 102 Q2: 700</em>

<em></em>

Similarly we find <em>m</em>, and <em>b</em>

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\\

<em>And we can write our Supply equation S(q):</em>

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

(c) Now we may find the equilibrium quantity q* and the equilibrium price p* by writing <em>D(q)=S(q)</em>, which means the demand <u><em>equals</em></u> 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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Answer:

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4 years ago
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