Use synthetic substitution to find |(3) for f(x) = x2

Precal. Find the sum. Show every step you took to get to your answer. I will give brainliest if you are correct and show all wor

Ymorist [56]

Answer:

I think your forgot to add some parts of this question, because you can't do anything with this except simplify it down a little maybe. But, I will provide the simplified version of this.

$\frac{x-4}{x^2-2x}+\frac{4}{x^2-4}$

= $\frac{x-4}{x\left(x-2\right)}+\frac{4}{x^2-4}$

= $\frac{x-4}{x\left(x-2\right)}+\frac{4}{\left(x+2\right)\left(x-2\right)}$

= $\frac{\left(x-4\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{4x}{\left(x+2\right)\left(x-2\right)x}$

= $\frac{\left(x-4\right)\left(x+2\right)+4x}{x\left(x-2\right)\left(x+2\right)}$

= $\frac{x+4}{x\left(x+2\right)}$

Step-by-step explanation:

Hope this helped!

8 0

3 years ago

the total attendance at a concert in a theater hall was 1500, of this total 400 were children,850 were women , and the remaining

AleksandrR [38]

Answer:

Below!

Step-by step explanation:

We know that:

400 + 850 + m = 1500

Solution of Question A:

Percent of children: Total children/Total attendance

=> 400/1500
=> 4/15
=> 0.27 (Rounded to nearest hundredth)
=> 0.27 x 100
=> 27%

Hence, the percent of children is about 27%.

Solution of Question B:

Percent of women: Total women/Total attendance

=> 850/1500
=> 85/150
=> 17/30
=> 17/30 x 100
=> 17/3 x 10
=> 170/3
=> 56.67%

Hence, the percent of women is 56.67%.

Solution of Question C:

400 + 850 + m = 1500
=> 1250 + m = 1500
=> m = 1500 - 1250
=> m = 250

Percent of men: Total men/Total attendance

=> 250/1500
=> 1/6
=> 0.17 (Rounded to nearest hundredth)
=> 0.17 x 100
=> 17%

Hence, the percent of men is about 17%

Hoped this helped.

$BrainiacUser1357$

3 0

2 years ago

A fast food restaurant executive wishes to know how many fast food meals teenagers eat each week. They want to construct a 99% c

Lana71 [14]

Answer:

The minimum sample size required to create the specified confidence interval is 2229.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

What is the minimum sample size required to create the specified confidence interval

This is n when $M = 0.06, \sigma = 1.1$

$0.06 = 2.575*\frac{1.1}{\sqrt{n}}$

$0.06\sqrt{n} = 2.575*1.1$

$\sqrt{n} = \frac{2.575*1.1}{0.06}$

$(\sqrt{n})^{2} = (\frac{2.575*1.1}{0.06})^{2}$

$n = 2228.6$

Rounding up

The minimum sample size required to create the specified confidence interval is 2229.

7 0

3 years ago

Find the missing side length of the triangle 48 units 14 units

Tpy6a [65]

Answer:

what are the options? I would say something like 50-60

5 0

3 years ago

Of the 20 students in Emily's class 15 like chocolate cake and the rest like cheesecake how many like cheesecake and what percen

alisha [4.7K]

5 students like cheesecake because if there is 20 students in all and 15 like chocolate and the rest like cheesecake you do 20-15=5

If 5 students like cheesecake out of 20 it would be 5/20 20*5=100 just multiply by 5 5*5=25%

25% of students like cheesecake

4 0

3 years ago

Read 2 more answers

Use synthetic substitution to find |(3) for f(x) = x2 - 9x +5.