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2 years ago

5/9 rounded to the nearest half?

Mathematics

1 answer:

Archy [21]2 years ago

6 0

0.6.

5/9 = 0.5555555... which rounds to 0.6

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On a planet far far away from Earth, IQ of the ruling species is normally distributed with a mean of 106 and a standard deviatio

DENIUS [597]

Answer:

Step-by-step explanation:

N(106,14²)

(calculated using excel, let me know if you were supposed to use a normal table) = 0.4715

85.34

Q1:96.56

Q3:115.44

IQR (which is Q3-Q1)= 115.44-96.56= 18.88

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2 years ago

4. I NEED HELP ASAP!! PLEASE PUT AND ANSWER AND STEP BY STEP EQUATION!! IF YOU DON'T KNOW THE ANSWER DON'T PUT ANYTHING!!

tino4ka555 [31]

Answer:

The answer is A (y = 2x - 2)

Step-by-step explanation:

B and D aren't correct because the y-intercept is a positive 2 instead of negative 2 like the graph shown. We're now left with A and C. Go on one point of the graph and count up or down to go on the same y value as the other point. Go left or right and count until you go to the same point. Remember that the slope equation is rise/run. Rise meaning up or down. Run meaning left or right.

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2 years ago

Read 2 more answers

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}}\\

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

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\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}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{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 too easy 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.

Differential (Power Rules)

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

Differentiation (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)}}\\

\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)}}$

Differential (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}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\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}}\\

\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

2 years ago

Which real value of m makes the expression undefined M^8+1/4m^2

Soloha48 [4]

Answer:

m = 0

Step-by-step explanation:

The denominator of the rational expression cannot be zero as this would make the rational expression undefined. Equating the denominator to zero and solving gives the value that m cannot be.

4m² = 0 ⇒ m² = 0 ⇒ m = 0 ← excluded value

5 0

2 years ago

Helppppppppppppppppppppppppppppp

GalinKa [24]

Answer:
x = 4

Explanation:
For the right hand side:
log base 9 of 8 + log base 9 of (x-2) = log base 9 (8*(x-2))
= log base 9 of (8x-16)
For the left hand side:
2*log base 9 of x = log base 9 of x^2

Now, equating the two sides and solving for x:
log base 9 of (8x-16) = log base 9 of x^2
This means that:
x^2 = 8x - 16
x^2 - 8x + 16 = 0
(x-4)^2 = 0
Therefore:
x = 4

Hope this helps :)

7 0

2 years ago