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grigory [225]
3 years ago
9

The slope of the line below is 2. Use the coordinates of the labeled point to find a point-slope equation of the line.

Mathematics
1 answer:
givi [52]3 years ago
3 0

Answer:

1,9

Step-by-step explanation:

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Pleaseee helpp meeee
Marta_Voda [28]

Answer and Step-by-step explanation:

[] means to take the absolute version of the number. That means if the number is negative or positive, the outcome is always positive.

14.

-[3] = -3

15.

[5 + 9] = 14

16.

[17 - 8] = 9

#teamtrees #WAP (Water And Plant)

6 0
3 years ago
To estimate μ, the mean salary of full professors at American colleges and universities, you obtain the salaries of a random sam
GenaCL600 [577]

Answer:

<h3>73220±566.72</h3>

Step-by-step explanation:

The formula for calculating the confidence interval is expressed as;

CI = xbar ± z*s/√n

xbar is the sample mean =  $73,220

z is the z score at 99% CI = 2.576

s is the standard deviation = $4400

n is the sample size = 400

Substitute the given values into the formula;

CI = 73,220 ± 2.576*4400/√400

CI = 73,220 ± 2.576*4400/20

CI = 73,220± (2.576*220)

CI = 73220±566.72

Hence a 99% confidence interval for μ is 73220±566.72

5 0
3 years ago
How do you find the limit?
coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}

We can cancel x-5.

\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}

Then directly substitute x = 5 in.

\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\&#10;&#10;\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\&#10;&#10;\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}

Differentiate numerator and denominator, apply power rules.

<u>Differential</u> (Power Rules)

\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}

<u>Differentiation</u> (Property of Addition/Subtraction)

\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}

Hence from the expressions,

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}

<u>Differential</u> (Constant)

\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \  is \ \ a \ \ constant.)}}

Therefore,

\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}

Now we can substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}

Thus, the limit value is 2/5 same as the first method.

Notes:

  • If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.
8 0
3 years ago
Of every 5 CDs produced in a
Anestetic [448]

Answer:

880

Step-by-step explanation:

2/5 times 2200

4 0
3 years ago
Read 2 more answers
Roy Gross is considering an investment that pays 7.6 percent, compounded annually. How much will he have to invest today so that
Debora [2.8K]

Answer: he should invest $16129 today.

Step-by-step explanation:

Let $P represent the initial amount that should be invested today. It means that principal,

P = $P

It would be compounded annually. This means that it would be compounded once in a year. So

n = 1

The rate at which the principal would be compounded is 7.6%. So

r = 7.6/100 = 0.076

The duration of the investment would be 6 years. So

t = 6

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount in the account at the end of t years.

A = 25000

Therefore

25000 = P(1+0.076/1)^1×6

25000 = P(1.076)^6

25000 = 1.55P

P = 25000/1.55

P = $16129

4 0
3 years ago
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