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3 years ago

Find the quotient. (6x 2 + 23x + 20) ÷ (5 + 2x)

Mathematics

2 answers:

arsen [322]3 years ago

8 0

Answer:

The quotient is 3x+4

Step-by-step explanation:

Given the dividend and divisor i.e we have to find the quotient if $6x^2+23x+20$ is divided by 5+2x

Given dividend is

$6x^2+23x+20$

⇒ $6x^2+15x+8x+20$

⇒ $3x(2x+5)+4(2x+5)$

⇒ $(3x+4)(2x+5)$

Hence, the given dividend can written in two factors.

Now, $\frac{6x^2+23x+20}{5+2x}=\frac{(3x+4)(2x+5)}{2x+5}=3x+4$

Hence, the quotient is 3x+4

tekilochka [14]3 years ago

4 0

(3x -4)(2x-5) as this it is divided

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2 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 that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

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

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

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

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

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

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2 years ago

I need help finding the answer!!

Whitepunk [10]

Answer:

The second option

Step-by-step explanation:

Here, we need to multiply each part of the matrix by -10.

This will give us: [ -230 380 ]

[ -170 60 ]

So, the answer is the second option.

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