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Savatey [412]
2 years ago
3

The intersection of a right cylinder and a plane makes an oval cross section. Which option is true about the plane of intersecti

on?
The right cylinder is cut with a plane parallel to the base.

The right cylinder is cut with a plane perpendicular to the base and passing through the center.

The right cylinder is cut with a plane perpendicular to the base but not passing through the center.

The right cylinder is cut with a plane that is neither parallel nor perpendicular to the base.

Mathematics
2 answers:
dexar [7]2 years ago
8 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Among the choices provided above the answer should be <span>The right cylinder is cut with a plane that is neither parallel nor perpendicular to the base.</span>
Basile [38]2 years ago
6 0

Answer: The right cylinder is cut with a plane that is neither parallel nor perpendicular to the base.


Step-by-step explanation:

Given: The intersection of a right cylinder and a plane makes an oval cross section.

The only method to get an oval cross is "the right cylinder is cut with a plane that is neither parallel nor perpendicular to the base."

If the right cylinder is cut with a plane parallel to the base then shape of cross section will be circle.

If the right cylinder is cut with a plane perpendicular to the base and passing through the center then shape of cross section will be rectangle.

If the right cylinder is cut with a plane perpendicular to the base but not passing through the center then again shape of cross section will be rectangle.

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<u>Solution- </u>

A researcher placed a petri dish with 32,000 bacterial cells. One hour after being placed in the vacuum chamber, the number of cells in the petri dish had halved. Another hour later, the number of cells had again halved.

This can be represented as exponential decreasing function,

y=a(1 - r)^x

Where,

  • a = starting amount  = 32000
  • r = rate  = 50% = 0.5 as the sample becomes halved in each hour
  • x = hours

Putting the values,

\Rightarrow y=32000(1 - 0.5)^x

\Rightarrow y=32000(0.5)^x

y-intercept means, where x=0, so

\Rightarrow y=32000(0.5)^0\\\\\Rightarrow y=32000\times 1=32000

The coordinate of this poin will be (0, 32000)

This means when x=0 or at the starting of the research, the number of bacteria cells was 32000.

After 3 hours, number of bacteria cells will be,

\Rightarrow y=32000(0.5)^3

\Rightarrow y=32000\times 0.125

\Rightarrow y=4000

The  coordinate of this point will be (3, 4000)

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Which triangle’s unknown side length measures Start Root 53 End Root units?
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Answer:

47 is your answer

Step-by-step explanation:

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Milton spilled some ink on his homework paper. He can't read the coefficient of $x$, but he knows that the equation has two dist
Lera25 [3.4K]

Answer:

Sum = -81

Step-by-step explanation:

See the comment for complete question.

Given

c = 36 ----- Constant

No coefficient of x^2

Required:

Determine the sum of all distinct positive integers of the coefficient of x

Reading through the complete question, we can see that the question has 3 terms which are:

x^2 ---- with no coefficient

x ---- with an unknown coefficient

36 ---- constant

So, the equation can be represented as:

x^2 + ax + 36 = 0

Where a is the unknown coefficient

From the question, we understand that the equation has two negative integer solution. This can be represented as:

x = -\alpha and x = -\beta

Using the above roots, the equation can be represented as:

(x + \alpha)(x + \beta) = 0

Open brackets

x^2 + (\alpha + \beta)x + \alpha \beta = 0

To compare the above equation to x^2 + ax + 36 = 0, we have:

a = \alpha + \beta

\alpha \beta = 36

Where: \alpha, \beta and \alpha \ne \beta

The values of \alpha and \beta that satisfy \alpha \beta = 36 are:

\alpha = -1 and \beta = -36

\alpha = -2 and \beta = -18

\alpha = -3 and \beta = -12

\alpha = -4 and \beta = -9

So, the possible values of a are:

a = \alpha + \beta

When \alpha = -1 and \beta = -36

a = -1 - 36 = -37

When \alpha = -2 and \beta = -18

a = -2 - 18 = -20

When \alpha = -3 and \beta = -12

a = -3 - 12 = -15

When \alpha = -4 and \beta = -9

a = -4 - 9 = -13

At this point, we have established that the possible values of a are: -37, -20, -15 and -9.

The required sum is:

Sum = -37 -20 -15 - 9

Sum = -81

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