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3 years ago

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If u(x)=-2x^2+3 and v(x)=1/x what is the range of (u*v)(x) **the picture should help

2 answers:

nasty-shy [4]3 years ago

5 0

We have that
$u(x)=-2 x^{2} +3 \\ \\ v(x)= \frac{1}{x}$
$u(v(x))=u( \frac{1}{x})=-2( \frac{1}{x} )^{2} +3 \\ u( \frac{1}{x})= \frac{(3 x^{2} -2)}{ x^{2} }$

using a graph tool
see the attached figure

The horizontal asymptote of this function is at <span>y=3</span><span>.
So,
the range of this function is from </span><span>(−∞,3<span>)</span></span>

il63 [147K]3 years ago

5 0

Answer:

-OO,3

Step-by-step explanation:

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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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Answer:

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Step-by-step explanation:

Step 1: Write an equation

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Step 2: Use the Addition Property of Equality

2x = 16

Step 3: Use the Division Property of Equality

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