For a class activity, a teacher put students into teams of 5. The teams were made up of 2 students randomly chosen from group A and 3 students randomly chosen from group B.

If group A has 4 students and group B has 6 students, how many different team combinations were possible?

Enter your answer as a number, like this: 42

Answer:

Step-by-step explanation:

. Volume: A measure of _space  occupied by a _three dimensional _ figure.

1. Base: The surface on which an object stands on.

1. Height: The _vertical distance from top to bottom, creates a _90° degree angle with the base.

1. Inverse Operation: The opposite of a math operation; the opposite of addition is subtraction and the opposite of multiplication is division.

1. Diameter: A straight line going from one side of a point on a circle to the other through the _center.

1. Radius: The distance from the center to the point of a circle;or half of the diameter.

Volume of a Cylinder

A cylinder is a three dimensional object with a circular base and top.

To find the volume of a cylinder we use the following formula:πr²h

The sum we want is

[tex]\displaystyle \sum_{n=0}^\infty \frac{(-1)^{T_n}}{(2n+1)^2} = 1 - \frac1{3^2} - \frac1{5^2} + \frac1{7^2} + \cdots[/tex]

where [tex]T_n=\frac{n(n+1)}2[/tex] is the n-th triangular number, with a repeating sign pattern (+, -, -, +). We can rewrite this sum as

[tex]\displaystyle \sum_{k=0}^\infty \left(\frac1{(8k+1)^2} - \frac1{(8k+3)^2} - \frac1{(8k+7)^2} + \frac1{(8k+7)^2}\right)[/tex]

For convenience, I'll use the abbreviations

[tex]S_m = \displaystyle \sum_{k=0}^\infty \frac1{(8k+m)^2}[/tex]

[tex]{S_m}' = \displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{(8k+m)^2}[/tex]

for m ∈ {1, 2, 3, …, 7}, as well as the well-known series

[tex]\displaystyle \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}[/tex]

We want to find [tex]S_1-S_3-S_5+S_7[/tex].

Consider the periodic function [tex]f(x) = \left(x-\frac12\right)^2[/tex] on the interval [0, 1], which has the Fourier expansion

[tex]f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi nx)}{n^2}[/tex]

That is, since f(x) is even,

[tex]f(x) = a_0 + \displaystyle \sum_{n=1}^\infty a_n \cos(2\pi nx)[/tex]

where

[tex]a_0 = \displaystyle \int_0^1 f(x) \, dx = \frac1{12}[/tex]

[tex]a_n = \displaystyle 2 \int_0^1 f(x) \cos(2\pi nx) \, dx = \frac1{n^2\pi^2}[/tex]

(See attached for a plot of f(x) along with its Fourier expansion up to order n = 10.)

Expand the Fourier series to get sums resembling the [tex]S'[/tex]-s :

[tex]\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{k=0}^\infty \frac{\cos(2\pi(8k+1) x)}{(8k+1)^2} + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+2) x)}{(8k+2)^2} + \cdots \right. \\ \,\,\,\, \left. + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+7) x)}{(8k+7)^2} + \sum_{k=1}^\infty \frac{\cos(2\pi(8k) x)}{(8k)^2}\right)[/tex]

which reduces to the identity

[tex]\pi^2\left(\left(x-\dfrac12\right)^2-\dfrac{21}{256}\right) = \\\\ \cos(2\pi x) {S_1}' + \cos(4\pi x) {S_2}' + \cos(6\pi x) {S_3}' + \cos(8\pi x) {S_4}' \\\\ \,\,\,\, + \cos(10\pi x) {S_5}' + \cos(12\pi x) {S_6}' + \cos(14\pi x) {S_7}'[/tex]

Evaluating both sides at x for x ∈ {1/8, 3/8, 5/8, 7/8} and solving the system of equations yields the dependent solution

[tex]\begin{cases}{S_4}' = \dfrac{\pi^2}{256} \\\\ {S_1}' - {S_3}' - {S_5}' + {S_7}' = \dfrac{\pi^2}{8\sqrt 2}\end{cases}[/tex]

It turns out that

[tex]{S_1}' - {S_3}' - {S_5}' + {S_7}' = S_1 - S_3 - S_5 + S_7[/tex]

so we're done, and the sum's value is [tex]\boxed{\dfrac{\pi^2}{8\sqrt2}}[/tex].

Answer:

sin C = 3/5

Step-by-step explanation:

see image.

It helps to draw a picture. Tan C is the ratio of the OPP/ADJ.

Pythagorean theorem or if you know Pythagorean triples are a shortcut to find the hypotenuse.

Once you know the hypotenuse, use the ratio for sine to solve the question. Sine is OPP/HYP.

see image.

Answer:

See below, please

Step-by-step explanation:

[tex](2x + 9) + (4x - 3) = 90[/tex]

[tex]6x + 6 = 90[/tex]

[tex]6x = 90 - 6 = 84[/tex]

Hence

[tex]x = 14[/tex]

6 red face cards

->in favour:

6/52

= 3/26

-> against:

52-6= 46

46/52

=23/26

With the frequecy distribution shown in the 50 cities of pakistan,

= highest value - lowest value

= 26.5 - 8.5

= 18

= ∑ f x / ∑ f

= ∑ f x / N

= 905 / 50

= 18.1

median

= lower limit + ( N/2 - C ) * h / ( frequency of the class interval )

C = cumulative frequency preceeding to the median class frequency

h = class interval

= 18.5 + ( 50 / 2 - ( 5 + 12 ) ) * 3 / 18

= 18.5 + 1.3333

= 19.8333

The mode is the value with the highest frequency occurence. This is under class 18 - 20

mode = lower limit + ( ( f1 - f0 ) / (2*f1 - f0 - f2 ) ) * h

f1 = fequency of the modal class

f0 = freqency of the preceeding modal class

f2 = frequency of the next modal class

h = class interval

= 18.5 + ( ( 18 - 12 ) / (2 * 18 - 12 - 8 )  ) * 3

= 18 + ( 0.375 ) * 3

= 19.125

= sqrt ( 1 / N ∑ f ( x - x' )^2 )

= sqrt (1  / 50 * 706.5

= 3.7589

=  standard deviation / mean

= 3.7589 / 3

= 1.2530

= ( standard deviation )^2

= ( 3.7589 )^2

= 14.1293

=  ∑ f ( x - x' )^4 / ( N * ( standard deviation )^4 )

= 27459.405 / ( 50 * 3.7589^4 )

= 2.7509

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[tex]\sqrt{7^{16}} = 7^n\\\\\implies \left(7^{16}\right)^{\tfrac 12} = 7^n\\\\\implies 7^{\left(\tfrac 12 \times 16\right)}=7^n\\\\\implies 7^8 = 7^n\\\\\implies \ln 7^8 = \ln 7^n\\\\\implies 8\ln 7 = n \ln 7\\\\\implies n =8[/tex]

Fort nite battle pass is 8 dollars

Answer:

Step-by-step explanation:

[tex]7 - 1 \frac{3}{5} = 6 \frac{2}{5}[/tex]

Answer:

688.19 inches

Step-by-step explanation:

Answer: C

Step-by-step explanation:

Just look at a z-score table and multiply by 100.

-> (0.308538)(100) is about 30.85%

Answer:

226

Step-by-step explanation:

Given:

Simplify 6^3+5(4-2)

Note:

I think you meant 6^3 because if you solve 63+5(4-2):

63+5(4-2)

63+5 * 2

63 + 10

73
Solve:

6^3 + 5(4 - 2 )    

6^3 + 5 x 2

6 x 6 x 6 = 216

226 + 5 x 2

5 x 2 = 10

216 + 10 = 226

~Lenvy~

The fact that the angles in a triangle add up to 180 indicates that it is actually a square). There are also four right triangles, each with a base and a height. The Pythagorean Theorem is reached when a2 + b2 = c2.

The function L= 10 log(I/10^-16) is a logarithmic equation

The loudness of the whisper is 40 decibels

The function of the loudness is given as:

L= 10 log(I/10^-16)

When the intensity is 10^-12, the equation becomes

L= 10 log(10^-12/10^-16)

Evaluate the quotient

L= 10 log(10^4)

Apply the rule of logarithm

L= 10 * 4

Evaluate the product

L = 40

Hence, the loudness of the whisper is 40 decibels

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Answer:

  840 km

Step-by-step explanation:

The speed expression ...

  840 kilometers per hour

means the plane files 840 kilometers in each hour.

In 1 hour, it will travel 840 km.

k must be greater than or equal to 22.75 to have two different zeros.

Second order polynomials are algebraic expressions that observe the following form:

[tex]p(x) = a\cdot x^2 + b\cdot x + c[/tex]   (1)

Where:

For polynomials of the form p(x) = 0, we can infer the nature of their roots by applying the following discriminant:

d = b² - 4 · a · c   (2)

According to (2), there are three cases:

Now we have the following discriminant case:

-(3 + 2 · k)² - 4 · (1) · (4) ≠ 0

-(9 + 6 · k + 4 · k²) - 16 ≠ 0

-9 - 6 · k - 4 · k² - 16 ≠ 0

4 · k²+ 6 · k +25 ≠ 0

This characteristic polynomial has two conjugated complex roots, then we conclude that all values of k must positive or negative, but never zero. By graphng tools we find that k must be greater than or equal to 22.75 to have two different zeros.

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Answer:

The answer is c on edge or f(x) = 1 sqt x + 6 -2

Step-by-step explanation:

From the given two options, none of them has a function that has the same maximum value as f(x) = -|x+3|-2.

A function is a correspondence between input numbers (x-values) and output numbers (y-values). It is used to describe an equation.

Given that:

Suppose that x = c is a critical point of (x) then,

If f'(x) > 0 to the left of x = c and f'(x) < 0 to the right of x = c;

If f'(x) < 0 to the left of x = c and f'(x) > 0 to the right of x = c;

If f'(x) is the same sign on both sides of x = c;

From the given equation, the critical points: x = -3

If we put the point x = -3 into - |x+3|-2

We can therefore conclude that none of them has a function that has the same maximum value as f(x) = -|x+3|-2.

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#SPJ9

Answer:

4

Step-by-step explanation:

100 divided by 25 equals 4.

Answer:

[tex]\frac{dx^{2} (x+1)S^{2} }{2(x^{2} +6x+3)^{2} }+ C[/tex]

Step-by-step explanation:

Proof -

So, in the first part we'll verify by taking n = 1.

[tex] \implies \: 1 = {1}^{2} = \frac{1(1 + 1)(2 + 1)}{6} [/tex]

[tex] \implies{ \frac{1(2)(3)}{6} }[/tex]

[tex]\implies{ 1}[/tex]

Therefore, it is true for the first part.

In the second part we will assume that,

[tex] \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} = \frac{k(k + 1)(2k + 1)}{6} }[/tex]

and we will prove that,

[tex]\sf{ \: { {1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}[/tex]

[tex] \: {{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k + 1)(k + 2) (2k + 3)}{6}}[/tex]

[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} + \frac{(k + 1) ^{2} }{6} [/tex]

[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{k(k+1)(2k+1)+6(k+1)^ 2 }{6} [/tex]

[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)\{k(2k+1)+6(k+1)\} }{6}[/tex]

[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2 +k+6k+6) }{6} [/tex]

[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6} [/tex]

[tex]{1}^{2} + {2}^{2} + {3}^{2} + ..... + {k}^{2} + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6} [/tex]

Henceforth, by using the principle of mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n.

_______________________________

Please scroll left - right to view the full solution.

Answer:

Because 10 is twice as large as 5.

Step-by-step explanation:

missing angle:

180° - 90° - 30°

180° - 120°

60°

missing sides:

(a)

[tex]\rightarrow \sf tan(x)= \dfrac{opposite}{adjacent}[/tex]

[tex]\rightarrow \sf tan(30)= \dfrac{4}{adjacent}[/tex]

[tex]\rightarrow \sf adjacent= \dfrac{4}{tan(30)}[/tex]

[tex]\rightarrow \sf adjacent= 4\sqrt{3}[/tex]

[tex]\rightarrow \sf adjacent= 6.93 \ cm[/tex]

(b)

[tex]\sf \rightarrow sin(x)= \dfrac{opposite}{hypotensue}[/tex]

[tex]\sf \rightarrow sin(30)= \dfrac{4}{hypotensue}[/tex]

[tex]\sf \rightarrow hypotensue= \dfrac{4}{ sin(30)}[/tex]

[tex]\sf \rightarrow hypotensue= 8 \ cm[/tex]

Answer:

m∠X = 60°

BX = 8 cm

BM = 4√3 cm

Step-by-step explanation:

The sum of the interior angles of a triangle is 180°

Given:

⇒ m∠B + m∠M + m∠X = 180°

⇒ 30° + 90° + m∠X = 180°

⇒ 120° + m∠X = 180°

⇒  m∠X = 180° - 120°

⇒  m∠X = 60°

Using the sine rule to find the side lengths:

[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

(where A, B and C are the angles, and a, b and c are the sides opposites the angles)

Given:

[tex]\implies \dfrac{4}{\sin 30\textdegree}=\dfrac{BX}{\sin 90\textdegree}=\dfrac{BM}{\sin 60\textdegree}[/tex]

[tex]\implies BX=\sin 90\textdegree \cdot\dfrac{4}{\sin 30\textdegree}[/tex]

              [tex]=1 \cdot \dfrac{4}{\frac12}[/tex]

              [tex]=1 \cdot 4 \cdot 2[/tex]

              [tex]=8 \textsf{ cm}[/tex]

[tex]\implies BM=\sin 60\textdegree \cdot\dfrac{4}{\sin 30\textdegree}[/tex]

              [tex]=\dfrac{\sqrt{3}}{2}\cdot \dfrac{4}{\frac12}[/tex]

              [tex]=\dfrac{\sqrt{3}}{2}\cdot 4 \cdot 2[/tex]

              [tex]=4\sqrt{3} \textsf{ cm}[/tex]

Answer:

(0,-9) You have to substitute 0 for x and solve for y

Answer:

Rs. 32767

Step-by-step explanation:

Because the amount is doubling every day, we can use the expression 1*2^15-1 because there is 1 to start with. Also cool trick! if you need to do 2^1+2^2+2^3+....+2^x, it will be equal to 2^(x+1)-1. So:

2^15-1

32768-1

32767

Answer:

Answer:4 apples weigh 5/8 pound.

Step-by-step explanation:

Answer:

2(−5) − 10 = 2(0)

Step-by-step explanation:

If you substitute the values x = 0 and y = −5 into the second equation, you get a false statement

The probability of not drawing C in neither draw is P = 0.5

All the cards have the same probability of being drawn, in this case, our set of cards is {F, D, C, G}

The probability of not drawing C is equal to the probability of drawing F, D or G. So we have 3 options out of 4, then the probability is:

p = 3/4.

Now we draw another, this time there are 3 cards, one of these is C, and the other two cards are not C. Then the probability of not drawing C again is equal to 2 over 3.

q = 2/3.

The joint probability (for both of these events to happen) is equal to the product of the individual probabilities:

P = p*q = (3/4)*(2/3) = 0.5

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