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Yuri [45]
2 years ago
9

The circumference of a circle is approximately 43.96 centimeters. What is the radius of the circle? 3.14 for Pie

Mathematics
1 answer:
Scilla [17]2 years ago
3 0

Answer:

7 Centimeters

Step-by-step explanation:

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Can 40:70 be simplified?
Veseljchak [2.6K]
No they are not equivalent and if you are trying to get a mixed number try dividing
8 0
3 years ago
6n+7n<br> whats the answer
sergeinik [125]
13n





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3 0
3 years ago
Read 2 more answers
Solve the following questions-
Setler79 [48]

Answer:

  2/5

Step-by-step explanation:

To get the ratio as a pure number, it must be expressed as the quotient of two values that have the same units. For the purpose here, it is convenient to convert both values to units of seconds.

__

<h3>units conversion</h3>

The conversion factor between minutes and seconds is ...

  1 minute = 60 seconds

Multiplying this equation by 3 gives ...

  3 minutes = 180 seconds

__

<h3>ratio of interest</h3>

Then the desired ratio is ...

  (72 seconds)/(3 minutes) = (72 seconds)/(180 seconds) = 72/180

  = (36×2)/(36×5)

  = 2/5

The ratio in its simplest form is 2/5.

3 0
2 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
^^^^<br> please help :)))
ikadub [295]

2x-1.1/4+x/3=2

2x-1/4+x/3=2

2x-1/4+x/3+1/4=2+1/4

2x+x/3=9/4

3(2x+x/3)=3.94

7x=27/4

7x/7=27/4/7

x=27/8

<h2>PLEASE MARK ME AS BRAINLIEST IF U LIKE MY ANSWER AND SORRY FOR GIVING THE ANSWER LATE BECAUSE I'VE GIVEN U ANSWER FROM MY LAPTOP PLEASE TELL THAT IT'S CORRECT OR NOT</h2>

7 0
3 years ago
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