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Natasha2012 [34]
3 years ago
13

3. (a) Are triangles PQR and STU congruent?

Mathematics
1 answer:
yawa3891 [41]3 years ago
8 0

We already know that PQ and ST are congruent because they both equal 4 and QR and TU are congruent because they both equal 6. Also I'm not sure if it is marking the two angles congruent but it looks like that to me so if that is the case then both of the triangles are congruent by SAS, PQ=ST, WR=TU, angle PQR = angle STU.

Now that we need to set up a system of equations to find y. Since we already proved the triangles congruent SU must be = to PR because of CPCTC "corresponding parts of congruent triangles are congruent" So the equation would be:

3y-2=y+4 (move y variable to one side) subtract y from right, add 2 to right

2y=6 (divide)

y=3

Then plug 3 into where y is (and since we need PQR perimeter plug it into 3(3)-2) and PR should equal 7 so add up all the sides (4+6+7) and your perimeter is 17ft.

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Step-by-step explanation:

Revenue (r)= p * n

where,

p = price per item

n = number of items sold

A change in price leads to a change in number sold

A variable to measure the change in p and n needs to be introduced

Let the variable=x

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p + x means a one dollar price increase

p - x means a one dollar price decrease

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n - x means a one item number-sold decrease

for each $2 price increase (p + 2x) there are 3 fewer rooms are rented (n-3x)

know that at $60 per room, the hotel rents 210 rooms

r = (60 + 2x) * (210 - 3x)

=12,600-180x+420x-6x^2

=12,600+240x-6x^2

r=2100+40x-x^2

= -x^2 +40x+2100=0

Solve the quadratic equation

x= -b +or- √b^2-4ac / 2a

a= -1

b=40

c=2100

x= -b +or- √b^2-4ac / 2a

= -40 +or- √(40)^2 - (4)(-1)(2100) / (2)(-1)

= -40 +or- √1600-(-8400) / -2

= -40 +or- √ 1600+8400 / -2

= -40 +or- √10,000 / -2

= -40 +or- 100 / -2

x= -40+100/-2 OR -40-100/-2

=60/-2 OR -140/-2

= -30 OR 70

x=70

The quadratic equation has a maximum at x=70

p+2x

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=60+60

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r= (60 + 2x) * (210 - 3x)

={60+2(30)}*{(210-3(30)}

r=(60+60)*(210-90)

=120*120

=$14,400

7 0
3 years ago
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