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tiny-mole [99]
3 years ago
9

(+5)+(-2)+(+20) solve

Mathematics
2 answers:
Veseljchak [2.6K]3 years ago
5 0

Answer:

5(x-2)^2=20

5(x^2-4)=20

5x^2-20=20

5x^2-20-20=0

5x^2-40=0

5x^2+0x-40=0

Step-by-step explanation:

hope this is helpfull

Darina [25.2K]3 years ago
4 0

Answer:

23

Step-by-step explanation:

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I will attach google sheet that I used to find regression equation.
We can see that linear fit does work, but the polynomial fit is much better.
We can see that R squared for polynomial fit is higher than R squared for the linear fit. This tells us that polynomials fit approximates our dataset better.
This is the polynomial fit equation:
T(h)=20.2-3.6h-0.875h^2
I used h to denote hours. Our prediction of temperature for the sixth hour would be:
T(6)=20.2-3.6(6)-0.875(6)^2=-32.9
Here is a link to the spreadsheet  (<span>https://docs.google.com/spreadsheets/d/17awPz5U8Kr-ZnAAtastV-bnvoKG5zZyL3rRFC9JqVjM/edit?usp=sharing)</span>

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3 years ago
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charlie has 2 fair coins. he repeatedly tosses the pair of coins simultaneously (i.e., two tosses at a time), until he has seen
Serggg [28]

By using the trial method we get a total number of trials taken by Charlie to see both sides of both the coins is 4.

<h3>What is probability?</h3>

Probability is the name of the area of mathematics that deals with the examination of random events. The ratio of favorable occurrences to the total number of events is used to calculate an event's probability.

P(E) = F(E)/T (E)P(E)

It stands for the probability that an event will occur.

F(E) = Amount of favorable occurrences

Total number of trials (T(E))

Given that Charlie has 2 fair coins.

If he tosses the pair of coins simultaneously, then the number of samples can be HH, HT, TH, TT.

So to see both sides of both the coins he should toss the coin four times.

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7 0
1 year 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.
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gizmo_the_mogwai [7]

Answer:

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Step-by-step explanation:

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