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Evgesh-ka [11]
3 years ago
5

Combinations- how to show that 10C2=10C8

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
7 0
By definition,

{}^nC_k=\dbinom nk=\dfrac{n!}{k!(n-k)!}

We have

\dbinom{10}2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10!}{(10-8)!8!}=\dbinom{10}8

and so

{}^{10}C_8={}^{10}C_2
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Find the slope of the line passing through the points (-9, 3) and (7, -7).
Viktor [21]

Answer:

-10/16

Step-by-step explanation:

-7z3=-10 and 7-(-9)=16

6 0
2 years ago
The data set is 7,8,10,11. How would the mean, median, and mode change if you added a 9 to the data set?
andrezito [222]
The mean would stay the same, the median would become 9, and the mode would stay the same.
3 0
3 years ago
How are<br> negative<br> exponents and<br> positive<br> exponents<br> related?
Stolb23 [73]

You can make a negative exponent positive by switching its numerator with its denominator and vice versa.

Examples:

x^{-2} = \frac{1}{x^2}

3x^{-2} = \frac{3}{x^2}

(3x)^{-2} = \frac{1}{(3x)^2}

\frac{1}{x^{-2}} = x^2

\frac{1}{3x^{-2}} = \frac{x^2}{3}

\frac{1}{(3x)^{-2}} = \frac{(3x)^2}{1} = (3x)^2

7 0
3 years ago
Will crown the correct answers brainliest 3x
Solnce55 [7]

Answer :

(1) \frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}

(2) \frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{15xy}

(3) \frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{14xy}

(4) \frac{2}{5x}+\frac{3}{7y}=\frac{6y+15x}{21xy}

(5) \frac{7}{11x}-\frac{1}{33y}=\frac{231y-11x}{363xy}

(6) \frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{7y+42y-2x}{14xy}

Step-by-step explanation :

(1) The given expression is: \frac{1}{x}+\frac{1}{y}

\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}

(2) The given expression is: \frac{1}{5x}+\frac{1}{3y}

\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{(5x)\times (3y)}=\frac{3y+5x}{15xy}

(3) The given expression is: \frac{1}{7x}-\frac{1}{2y}

\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{(7x)\times (2y)}=\frac{2y-7x}{14xy}

(4) The given expression is: \frac{2}{5x}+\frac{3}{7y}

\frac{2}{5x}+\frac{3}{7y}=\frac{(2\times 3y)+(3\times 5x)}{(5x)\times (7y)}=\frac{6y+15x}{21xy}

(5) The given expression is: \frac{7}{11x}-\frac{1}{33y}

\frac{7}{11x}-\frac{1}{33y}=\frac{(7\times 33y)-(1\times 11x)}{(11x)\times (33y)}=\frac{231y-11x}{363xy}

(6) The given expression is: \frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}

\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{(x\times 7y)+(3\times 2x\times 7y)-(2x\times x)}{(2x)\times (x)\times (7y)}=\frac{7xy+42xy-2x^2}{14x^2y}=\frac{7y+42y-2x}{14xy}

7 0
3 years ago
Find the ordered pairs for the x- and y-intercepts of the equation 5x-6y=30 and select the appropriate option below
Maurinko [17]

Answer:

Y=-5

X=6

Step-by-step explanation:

set x =0 then solve then set y=o and solve

6 0
3 years ago
Read 2 more answers
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