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Sholpan [36]
4 years ago
14

Name all the possible shapes the cross section can be for a cone?

Mathematics
1 answer:
rusak2 [61]4 years ago
4 0
Circle, Ellipse, Parabola, Hyperbola. Hope this helps
You might be interested in
Help me please please
gavmur [86]

Answer:

22. 10^10

8 Billion has 9 zeros and 10 to the power 10 means there is a total of 10 zeros. So 10 zeros is more closer to 9 zeros than that of 11 zeros.

23.Exponential form is when a certain number is raised to the power of a certain number or also known as exponents. Exponents signifies that the base or the number that is being raised to a certain powers will be multiplied a number of times, based on the exponents.

Exponential form is defined as repeated multiplication of  the base.

7 0
3 years ago
-15(y2)4 divide 5y3y4
Nadya [2.5K]

Answer:

-3y

Step-by-step explanation:

-15(y2)4 ÷5y3y4

Rewrite the division as a fraction

<u>-15(y2)4</u>

 5y3y4

Multiply y3 by y4 by adding the exponents

<u>-5(y2)4</u>

  5y7

Cancel the common factor of -15 and 5

<u>-3(y2)4</u>

  y7

Multiply the exponents in (y2)4

<u>-3y8</u>

 y7

Cancel the common factor of y8 and y7

Factor y7 out of -3y8

<u>y7(-3y)</u>

 y7

Multiply by 1

<u>y7(-3y)</u>

 y7 × 1

Cancel the common factor which is y7 and y7

Rewrite the expression

<u>-3y</u>

 1

Divide -3y by 1

-3y

7 0
2 years ago
Suppose 150 customers of a restaurant are chosen for a sample, but only 30 respond. What is this an example of?
ZanzabumX [31]

Answer:

The correct answer is:

Nonresponse bias (C.)

Step-by-step explanation:

Nonresponse bias in sampling is one that arises due to the unwillingness or inability of subjects in a population to participate in a survey.

7 0
3 years ago
Linear Algebra: Permutation Matrices Let M be the matrix {{0,0,0,1,0}, {0,0,1,0,0}, {0,1,0,0,0}, {0,0,0,0,1}{1,0,0,0,0}}. What i
FromTheMoon [43]

Multiplying M by any matrix A would return new matrix, B, in which

• the 1st row of B is equal to the 4th row of A,

• the 2nd row of B is equal to the 3rd row of A,

• the 3rd row of B is equal to the 2nd row of A,

• the 4th row of B is equal to the 5th row of A, and

• the 5th row of B is equal to the 1st row of A.

The pattern here is

1 => 4 => 5 => 1

2 => 3 => 2

Let {4, 3, 2, 5, 1} denote the matrix M, where each number refers to the row of the identity matrix, I.

Using this notation, the pattern above gives

M² = {5, 2, 3, 1, 4}

M³ = {1, 3, 2, 4, 5}

M⁴ = {4, 2, 3, 5, 1}

M⁵ = {5, 3, 2, 1, 4}

M⁶ = {1, 2, 3, 4, 5}

so that <em>n</em> = 6.

(Notice that the first cycle has length 3 and the second one has length 2; the minimum <em>n</em> needed here is then LCM(2, 3) = 6.)

6 0
3 years ago
Rearrange the formula y=<br><br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%20%2B%202%7D%7Bx%20-%204%7D%20" id="TexFormula1
Sophie [7]

Answer:

Step-by-step explanation:

y =\frac{x+2}{x-4}\\\\y*(x-4)=x+2\\\\yx-4y = x+ 2\\\\yx-4y-x = 2\\\\yx -x  = 2 +4y\\\\x(y-1)=2+4y\\\\x=\frac{2+4y}{y-1}

5 0
3 years ago
Read 2 more answers
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