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hammer [34]
3 years ago
10

The ordered pairs shown below are on the same line. (6,4) (12.6) Which equation describes the relationship between the x-coordin

ates and y-coordinates in the ordered pairs?
a. y = 3x - 14
B.y = 3x - 6
C. y= 1/3x + 2​
Mathematics
1 answer:
Airida [17]3 years ago
4 0
..................................
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a pizza costs $11 and $1.50 per topping write a equation for cost in dollars and the nunber of toppings ​
romanna [79]

Answer:

y= 11 + 1.50x. (y is the price, x is number of toppings)

Step-by-step explanation:

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2 years ago
Read 2 more answers
Answer and working out please!
frosja888 [35]

Answer:

17.0710678119 and 2.92893218813

Step-by-step explanation:

Theoretically, the wire should come out to be x+y=80,

and the sum of the areas of the squares should be (x/4)^2+(y/4)^2=300.

x/4 and y/4 being the length of one side of the square, we would then square that to find the area. The total wire is all x+y=80. X being one part and Y being the other.

Theoretically, using systems. i found the answers: 68.2842712474619,11.715728752538098

Again, these numbers are very large, and I they actually do both add up to each amount basically perfectly. If your teacher is asking for a rounded answer, that would've been helpful to know. But again, theoretically, those are the answers.

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Weekly wages at a certain factory are
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The answer should be 2.35
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2 years ago
A text font fits 12 characters per inch. Using the same font, how many characters can be expected per yard of text?
Shkiper50 [21]
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Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and t
WINSTONCH [101]

Answer:

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.  

How long will it take for this population to grow to a hundred rodents? To a thousand rodents?

Step-by-step explanation:

Use the initial condition when dp/dt = 1, p = 10 to get k;

\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}

Seperate the differential equation and solve for the constant C.

\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2}  }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}

You have 100 rodents when:

100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months

You have 1000 rodents when:

1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months

7 0
2 years ago
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