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tester [92]
2 years ago
12

Which of the following equations would produce a parabola? Check all that apply. –5x^2 + 4x – 7y + 14 = 0 –x^2 + 6x + y2 – 8 = 0

9x2 – 12xy + 4y2 + 4x – y + 5 = 0 12x + 6y – 5y2 + 14 = 0 20x2 + 10xy – y2 + 3x – 6y + 5 = 0
Mathematics
2 answers:
omeli [17]2 years ago
4 0

Answer:

A,C,D

Step-by-step explanation:

Got 100% on assignment

olga2289 [7]2 years ago
3 0

Answer:

A= –5x2 + 4x – 7y + 14 = 0

C= 9x2 – 12xy + 4y2 + 4x – y + 5 = 0

D= 12x + 6y – 5y2 + 14 = 0

Step-by-step explanation:

I finished the assignment from Edge2020

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Which table could be used with Table A to verify that the function modeling Fahrenheit temperature, F(C), based on a given Celsi
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Answer:

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C -20 -15 -5

F(C)  -4 5 23

Step-by-step explanation:

3 0
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A cheetah runs at a rate of 75 miles per hour about how many leaders can it run in 5 seconds​
umka21 [38]

Answer:

Distance in meters D = 167.75 meters

Step-by-step explanation:

Distance covered in 5 seconds is given by:

Distance, D = Speed (S) X Time (T)

Where,

Distance = ?

Speed = 75mph

Time = 5s

Since we are calculating distance in meters, we have to convert 75miles per hour to meters per second

75mph = 75miles/1hr X 1km/0.621miles X 1000m/1km X 1hr/60mins X 1min/60s

            = (75X1000) / (0.621X60X60) meter per second

            = 75000/2235.6 meter per second

            = 33.55 meter per second

∴ Distance D = 33.55 meter per second X 5 seconds

                      = 167.75 meters

3 0
2 years ago
A cylinder of volume 10pi with a circular bottom and no top is constructed out of two different matierials. the cost of the bott
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6 0
3 years ago
Two litters of a particular rodent species have been born; first with two brownhaired and one gray-haired (litter 1) and the oth
maks197457 [2]

Answer:

The probability that the animal chosen is brown-haired is 0.6333.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> : a brown-haired rodent

<em>B</em> : Litter 1

The information provided is:

P (A|B) =\frac{2}{3}\\\\P(A|B^{c})=\frac{3}{5}

The probability of selecting any of the two litters is equal, i.e.

P(B)=P(B^{c})=\frac{1}{2}

According to the law of total probability:

P(X)=P(X|Y_{1})P(Y_{1})+P(X|Y_{2})P(Y_{2})+...+P(X|Y_{n})P(Y_{n})

Compute the total probability of event <em>A</em> as follows:

P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})

         =[\frac{2}{3}\times\frac{1}{2}]+[\frac{3}{5}\times\frac{1}{2}]\\\\=\frac{1}{3}+\frac{3}{10}\\\\=\frac{10+9}{30}\\\\=\frac{19}{30}\\\\=0.6333

Thus, the probability that the animal chosen is brown-haired is 0.6333.

5 0
3 years ago
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