Determine if the sides are greater than (>), less than (<), or equal (=) to each other.

If m(x)=x+5/x-1 and n(x)=x-3 which function has the same domain as (m•n)(x)

worty [1.4K]

Consider the functions m and n:

i) $m(x)=\frac{x+5}{x-1}$ for x≠1.

the domain of m is all the real numbers except 1.

ii) n(x)=x-3 is a polynomial function, because x-3 is a linear polynomial.

The domain of any polynomial function is all real numbers. To determine the Range of this function, let c be a value in the range:

x-3=c, then x=3+c.

so if we want the function to produce c, we just set x=3+c, check:

n(3+c)=(3+c)-3=c.

This means that the Range of n is all real numbers.

What this means, is that the input (x) of the function n(x) can be any real number, and the output (n(x)) can also be any number.

so let n(x)=a, consider m(a):

m(a)=(a+5)/(a-1), clearly a≠1, this means n(x)=x-3≠1, that is x≠-4

This means that the domain of the composed function, m(n(x)) = say f(x),

is all Real numbers except 4.

Check: letting x be 4, would mean n(4)=4-3=1

which would mean m(n(4))=m(1)=(1+5)/(1-1)=6/0, which makes no sense.

Answer: Any function with domain R-{4}, has the same domain of (mon)(x)

6 0

2 years ago

What is the value of 4*(-1+2-3+4-3+6-7+....+100) Show your work. [Pls help I am really drowning in work. 50 points + brainliest

KengaRu [80]

Answer:

200

Step-by-step explanation:

We have:

$4(-1+2-3+4-5+6-7...+100)$

We can rearrange the numbers to obtain:

$4((-1-3-5-7-...-99)+(2+4+6...+100))$

From the left, we can factor out a negative. So:

$4(-(1+3+5+7+...+99)+(2+4+6...+100))$

In other words, we want to find the sum of all the odd numbers from 1 to 99.

And the sum of all the even numbers from 2 to 100.

Let's do each one individually:

Odd Terms:

We have:

$(1+3+5+7+...+99)$

We can use the arithmetic series formula, where:

$S=\frac{k}{2}(a+x_k)$

Where k is the number of terms, a is the first term, and x_k is the last term.

Since it's all the odd numbers between 1 and 99, there are 50 terms.

Our first term is 1 and our last term is 99. So, the sum of all the odd terms are:

$S=\frac{50}{2}(1+99})$

Divide the fraction. Add within the parentheses:

$S=25(100)$

Multiply:

$S=2500$

So, the sum of all the odd terms is 2500.

Even Terms:

We have:

$(2+4+6+...+100)$

Again, we can use the above formula.

Our first term is 2, last term is 100. And since it's from 2-100, we have 50 even terms. So:

$S=\frac{50}{2}(2+100)$

Divide and add:

$S=25(102)$

Multiply:

$S=2550$

We originally had:

$4(-(1+3+5+7+...+99)+(2+4+6...+100))$

Substitute them for their respective sums:

$4(-(2500)+2550)$

Multiply:

$4(-2500+2550)$

Add:

$=4(50)$

Multiply:

$=200$

So, the sum of our sequence is 200.

And we're done!

Note: I just found a <em>way</em> easier way to do this. We have:

$4\cdot(-1+2-3+4-5+6-7+...+100)$

Let's group every two terms together. So:

$=4((-1+2)+(-3+4)+(-5+6)...+(-99+100))$

We can see that they each sum to 1:

$=4((1)+(1)+(1)+...+(1))$

Since there are 100 terms, we will have 50 pairs, so 50 times 1. So:

$=4(50)$

Multiply:

$=200$

Pick which one you want to use! I will suggest this one though...

Edit: Typo

8 0

3 years ago

You toss a fair coin 10000 times. what are the odds of obtaining more than 5100 tails, approximately?

ella [17]

This can be solved by using the normal approximation to the binomial distribution.
mean = np = 10.000 * 0.5 = 5,000
The standard deviation is given by:
$S.D.= \sqrt{npq} = \sqrt{5000\times0.5} =50$
$z=\frac{5100-5000}{50}=2$
The probability of obtaining more than 5100 tails is 0.0228 and the probability of obtaining fewer than 5100 tails is 0.9772.
The odds of obtaining more than 5100 tails is therefore:
0.0228:0.9772 = 1:42.86.

3 0

2 years ago

Really need help not good in geometry

Mandarinka [93]

Answer:18

Step-by-step explanation:It is the same shape just 3 times the size.

4 0

3 years ago

Write the equation for the graphed function. I worked out y = 5 - 4 (x + 4)^2 but that isn't right. Any help would be appreciate

alexira [117]

Vertex form for a parabola with vertex (p,q) is

$y = a(x-p)^2 + q$