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igor_vitrenko [27]
2 years ago
7

Prime number after 167, 118, 82, 57, 41,

Mathematics
1 answer:
Fiesta28 [93]2 years ago
8 0

The prime numbers after 167, 118, 82, 57, 41 are 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2.

Prime numbers are numbers that are divisible by 1 and itself.

This are number greater than one with only two factors which are 1 and itself.

In other words, this number, when they are divided by another number other

than 1 or itself , one will get a remainder that is not zero.

Example of prime numbers are 2, 5, 7, 11, 13 etc.

The prime numbers after  167, 118, 82, 57, 41 are as follows:

37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2.

learn more about prime number here: brainly.com/question/26013414

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find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0​
Citrus2011 [14]

Answer:

Radius: r =\frac{\sqrt {21}}{6}

Center = (-\frac{3}{2}, -\frac{2}{3})

Step-by-step explanation:

Given

9x^2 + 9y^2 + 27x + 12y + 19 = 0

Solving (a): The radius of the circle

First, we express the equation as:

(x - h)^2 + (y - k)^2 = r^2

Where

r = radius

(h,k) =center

So, we have:

9x^2 + 9y^2 + 27x + 12y + 19 = 0

Divide through by 9

x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0

Rewrite as:

x^2  + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}

Group the expression into 2

[x^2  + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}

[x^2  + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}

Next, we complete the square on each group.

For [x^2  + 3x]

1: Divide the coefficient\ of\ x\ by\ 2

2: Take the square\ of\ the\ division

3: Add this square\ to\ both\ sides\ of\ the\ equation.

So, we have:

[x^2  + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}

[x^2  + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2

Factorize

[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2

Apply the same to y

[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}

Add the fractions

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}

[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}

[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}

Recall that:

(x - h)^2 + (y - k)^2 = r^2

By comparison:

r^2 =\frac{7}{12}

Take square roots of both sides

r =\sqrt{\frac{7}{12}}

Split

r =\frac{\sqrt 7}{\sqrt 12}

Rationalize

r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}

r =\frac{\sqrt {84}}{12}

r =\frac{\sqrt {4*21}}{12}

r =\frac{2\sqrt {21}}{12}

r =\frac{\sqrt {21}}{6}

Solving (b): The center

Recall that:

(x - h)^2 + (y - k)^2 = r^2

Where

r = radius

(h,k) =center

From:

[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}

-h = \frac{3}{2} and -k = \frac{2}{3}

Solve for h and k

h = -\frac{3}{2} and k = -\frac{2}{3}

Hence, the center is:

Center = (-\frac{3}{2}, -\frac{2}{3})

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The boys drink 7/10 of a water bottle

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Svet_ta [14]

Answer: 3rd option

Step-by-step explanation: The answer would be the third option. Let's go through each one and explain why it's correct or incorrect. First of all, our function is y = x - 3 so we can plug numbers into our X and Y. If we follow the function y = x - 3, we have 8 = 10 - 3 which is incorrect for the first one so that can't be it. For the nest one, we have 13 = 10 - 3 which also does not make sense. For the third one, we have 7 = 10 - 3 which is correct. And the last one, we have 3 = 10 - 2 which is wrong. Therefore, the third option is the answer.

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