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4 years ago

What does the solution to this system mean in context of this problem ?

Mathematics

1 answer:

zhenek [66]4 years ago

3 0

The solution to the system (3,11) means that Alley Cat and Bob's bowling will cost the same if 3 games are played.....the cost will be $ 11.00

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What is the following sum???? Please help ASAP

Nadusha1986 [10]

Answer:

ung pangalawang tuldok po :)

7 0

3 years ago

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While training for a race, Andre's dad ran 12 miles in 75 minutes on a treadmill. If he runs at

zepelin [54]

Answer:

50 minutes

Step-by-step explanation:

12/75=8/x

12x=600

x=50

50 minutes for 8 miles

4 0

3 years ago

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Let the average number of vehicles arriving at the gate of an amusement park per minute be equal to k, and let the average numb

artcher [175]

Step-by-step explanation:

The given rational function defined by the equation:

$T(r)=\frac{(2r-k)}{(2r^2- 2kr)}$

where :k = average number of vehicles arriving at the gate per minute

r = average number of vehicles admitted by the park attendants

T = the average waiting time in minutes for each vehicle

a) k = 26 , r = ?, T = 30 seconds

T(r) = 30 seconds

$30=\frac{(2r-30)}{(2r^2- 2\times 30r)}$

$60r^2-1800r=2r-30$

$60r^2-1802r+30=0$

on solving:

r = 0.016657 , 30.017

r = 30.017 (accept, given that r > k )

Admittance rate r that is necessary to keep the average waiting time T for each vehicle to 30 sec is 30.017.

b) k = 5.3, r = ?, T = 30 seconds

T(r) = 30 seconds

$30=\frac{(2r-5.3)}{(2r^2- 2\times 5.3r)}$

$60r^2-318r=2r-5.3$

$60r^2-318r-2r+5.3=0$

$60r^2-320r+5.3=0$

r = 0.016657 , 5.31672

r = 5.31672 (accept, given that r > k )

5.31672 park attendants will be needed to keep the average wait to 30 seconds.

7 0

3 years ago

An open-top box with a square base is to be constructed from 120 square centimeters of material. What dimensions will produce a

Mama L [17]

Answer:

Maximum length and breadth of the box is 6.32 cm and the height of the box is 3.16 cm

Step-by-step explanation:

Surface area of box

$4xy+x^2=120\\\Rightarrow y=\dfrac{120-x^2}{4x}$

Volume of box is

$V=x^2y\\\Rightarrow V=x^2\times \dfrac{120-x^2}{4x}\\\Rightarrow V=\dfrac{120x-x^3}{4}\\\Rightarrow V=30x-\dfrac{1}{4}x^3$

Differentiating with respect to $x$

$\dfrac{dV}{dx}=30-\dfrac{3}{4}x^2$

Equating with zero

$30-\dfrac{3}{4}x^2=0\\\Rightarrow x^2=\dfrac{30\times 4}{3}\\\Rightarrow x^2=40\\\Rightarrow x=6.32$

Double derivative of the volume

$\dfrac{d^2V}{dx^2}=-\dfrac{6x}{4}=-\dfrac{6\times 6.32}{4}=-9.48$

So, the volume is maximum at $x=6.32$

$y=\dfrac{120-x^2}{4x}=\dfrac{120-40}{4\times \sqrt{40}}\\\Rightarrow y=3.16$

So, the maximum length and breadth of the box is 6.32 cm and the height of the box is 3.16 cm.

5 0

3 years ago

From 1960 to 1980 , the consumer price index (cpi) increased from 29.6 to 82.4. If a frozen chicken pie cost $0.28 in 1960 and t

mojhsa [17]

CPI in 1960 = 29.6

CPI in 1980 = 82.4

Growth (%) in CPI = $\frac{\text{CPI in 1980 - CPI in 1960}}{\text{CPI in 1960}}$ × 100

⇒ Growth (%) in CPI = $\frac{82.4-29.6}{29.6}$ × 100

⇒ Growth (%) in CPI = 178.3783.. %

⇒ Growth (%) in CPI = 178.4%

Now, we know that price of frozen chicken pie in 1960 = $0.28

Also, we know that over 1960 to 1980, frozen chicken pie price grew at the same rate as CPI

Hence, Price of Chicken Pie in 1980 = Price of Chicken Pie in 1960 × (1 + Growth in CPI)

⇒ Price of Chicken Pie in 1980 = 0.28 × (1 + 178.4%)

⇒ Price of Chicken Pie in 1980 = 0.28 + (0.28 × 178.4%)

⇒ Price of Chicken Pie in 1980 = 0.28 + 0.49952

⇒ Price of Chicken Pie in 1980 = 0.28 + 0.50

⇒ Price of Chicken Pie in 1980 = $0.78

Hence, price of chicken pie in 1980 would be ~$0.78

8 0

3 years ago

Read 2 more answers