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Ann [662]
3 years ago
5

Y=x+2 y+1= – 4x Is this ordered pair a solution of the system of equations? Help:(

Mathematics
1 answer:
Shkiper50 [21]3 years ago
3 0
  • y=x+2.....(1)
  • y+1= – 4x......(2)

Put value of y in Equation 1 in Equation 2.

\looparrowright \sf y + 1 =  - 4x

\\  \\

\looparrowright \sf (x + 2) + 1 =  - 4x

\\  \\

\looparrowright \sf x + 2 + 1 =  - 4x

\\  \\

\looparrowright \sf x +3=  - 4x

\\  \\

\looparrowright \sf x  + 4x=  - 3

\\  \\

\looparrowright \sf 5x=  - 3

\\  \\

\looparrowright \sf x=   \dfrac{ - 3}{5}

\\  \\

Let's find value of y :

\\  \\

\looparrowright \sf y=x+2

\\  \\

\looparrowright \sf y= \dfrac{ - 3}{5} +2

\\  \\

\looparrowright \sf y= \dfrac{ - 3 + 10}{5}

\\  \\

\looparrowright \sf y= \dfrac{7}{5}

Solution:

\looparrowright \sf  \bigg(  \dfrac{ - 3}{5} , \dfrac{7}{5}  \bigg)

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4 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
erma4kov [3.2K]

Answer:

\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

Step-by-step explanation:

Given

5 tuples implies that:

n = 5

(h,i,j,k,m) implies that:

r = 5

Required

How many 5-tuples of integers (h, i, j, k,m) are there such thatn\ge h\ge i\ge j\ge k\ge m\ge 1

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed:  This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

^{n}C_r => ^{n + 4}C_5

^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}

^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}

Expand the numerator

^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

<u><em>Solved</em></u>

6 0
3 years ago
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