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Alchen [17]
2 years ago
10

Twice the quotient of a number and -24 is -8 Find the number. The number is ________.

Mathematics
1 answer:
Klio2033 [76]2 years ago
4 0

Answer: The number should be 6

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PLEASEEEEE HELP ME IF YOU GET THIS QUESTION!!
EleoNora [17]

Answer:

one and a quarter mile

please can i have brainliest

5 0
2 years ago
If the equation below represents the total cost of 3 boxes of cereal and 2 loaves of bread is $15," then x represents
Liula [17]

Answer:

C. The cost of a box of cereal

Step-by-step explanation:

In the question, we are given the number of 3 boxes of cereal, 2 loaves of bread, and that the total cost is $15. We can see these numbers in the equation, and putting it into context, we see that the 15 is on the right side of the equation, which indicates cash. Therefore, the 3 and 2 are accounted for, but their prices are unknown. X is a placeholder for the unidentified cost of a box of cereal, and we know that it's the boxes of cereal and not the loaves of bread because there are 3 and not 2.

8 0
2 years ago
The Jayden family eats at a restaurant that is having a 15% discount promotion. Their meal
Schach [20]

Answer: $80.44

Step-by-step explanation:

First of all, find 20% of 78.86, which is approximately 15.77.

Then, add it to the cost of the meal.

Finally, discount 15% of 94.63, which is approximately $14.19, and then subtract it from $94.63, and there's your answer.

3 0
2 years ago
Find the
Mazyrski [523]

Answer:

54 cm

Step-by-step explanation:

The area (A) of an equilateral triangle is

A = \frac{s^2\sqrt{3} }{4}  ( s is the side length ) then

\frac{s^2\sqrt{3} }{4} = 81\sqrt{3} ( multiply both sides by 4 to clear the fraction )

s²\sqrt{3} = 324\sqrt{3} ( divide both sides by \sqrt{3} )

s² = 324 ( take the square root of both sides )

s = \sqrt{324} = 18

Thus

perimeter = 3 × 18 = 54 cm

3 0
3 years ago
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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