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Sav [38]
3 years ago
15

A cone is sliced so the cross section is parallel to its base.

Mathematics
2 answers:
julsineya [31]3 years ago
8 0

Answer:

Option 1 - Triangle

Step-by-step explanation:

Given :  A cone is sliced so the cross section is parallel to its base.

To find : What is the shape of the cross section?

Solution :

When a cone is cut in such a way that the cross-section is perpendicular to the base and it passes through the vertex.

The shape of cross section formed is a triangle.

Refer the attached figure below.

The figure thus formed after cutting is in the shape of a triangle.

Therefore, Option 1 is correct.

Alika [10]3 years ago
3 0
The shape of the cross-section formed is a triangle, when a cone is cut in such a way that the cross-section is perpendicular to the base and it passes through the vertex.
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When Jean was asked to say what the 3 in the fraction 2
JulijaS [17]

Fractions are quotients of two integers

A better way to say what the 3 in 2/3 represents is that, 3 is the denominator

<h3>How to determine what 3 represents</h3>

The fraction is given as: 2/3

A fraction is represented as: Numerator/ Denominator

By comparison;

  • Numerator = 2
  • Denominator = 3

Hence, a better way to say what the 3 in 2/3 represents is that, 3 is the denominator

Read more about fractions at:

brainly.com/question/11562149

7 0
2 years ago
Rad 18x^5y + Rad 32xy^3 - Rad 128xy
True [87]
Sqrt(18x^5y) + sqrt(32xy^3) - sqrt(128xy)
You may separate the roots to produce solvable square roots!

sqrt(9x^4)*sqrt(2xy) + sqrt(16y^2)*sqrt(2xy) - sqrt(64)*sqrt(2xy)

Simplifying makes it ...
(3x^2 + 4y - 8)*sqrt(2xy)
This should be the closest you can get to a simplified equation.
5 0
3 years ago
A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple
tester [92]

Answer:  The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation:  We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.

Therefore, the probability of event A is given by

P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

4 0
2 years ago
Who can answer this question
lyudmila [28]
The answer is: c
Explination 30-9=21
8 0
2 years ago
Trisha and Byron are washing and vacuuming cars to raise money for a class trip. Trisha raised $38 washington 5 cars and vacuumi
ANEK [815]

Answer:

They charged <u>$6</u> for washing and <u>$2</u> for vacuuming the car.

Step-by-step explanation:

Let the charge of washing a car be 'x'.

And also let the charge of vacuuming a car be 'y'.

Now according to question, Trisha raised $38 washing 5 cars and vacuuming 4 cars.

So framing the above sentence in equation form, we get;

5x+4y=38\ \ \ \ \ equation\ 1

Again according to question, Byron raises $28 by washing 4 cars and vacuuming 2 cars.

So framing the above sentence in equation form, we get;

4x+2y=28\ \ \ \ \ equation\2

Now multiplying equation 2 by '2', we get;

2(4x+2y)=28\times2\\\\8x+4y=56\ \ \ \ equation\ 3

Now we subtract equation 1 from equation 3 and get;

(8x+4y)-(5x+4y)=56-38\\\\8x+4y-5x-4y=18\\\\3x=18\\\\x=\frac{18}{3}=6

Now substituting the value of 'x' in equation 1, we get;

5x+4y=38\\\\5\times6+4y=38\\\\30+4y=38\\\\4y=38-30=8\\\\y=\frac{8}{4}=2

Hence They charged <u>$6</u> for washing and <u>$2</u> for vacuuming the car.

8 0
3 years ago
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