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erma4kov [3.2K]
2 years ago
11

Assume that 0 a. 16/63 b. -16/63 c. 63/16 d. -63/16 please help

Mathematics
1 answer:
iVinArrow [24]2 years ago
3 0
The answer to this question is B
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Bruce has a 16 ounces bottle of water that is 1/2 full. He drinks 1/5 of the water left in the bottle. How many ounces of water
Natali5045456 [20]

Answer: 1.6 Ounces.

16/2 is 8.

8/5 is 1.6

You can check this by multiplying 1.6 times 5 then by 2,

8 0
2 years ago
there are 342 seats in the Jefferson middle school auditorium. There are 75 seats being used by teachers. Students are using 2/3
V125BC [204]

Answer:

267 i think. I might have counted wrong



8 0
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 <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
2 years ago
What choices are equivalent to the quotient below ?
pshichka [43]

OPTION A , C,D

are the correct choices because all of them are equal to √5

4 0
2 years ago
4. The Bulldog Theater charges $9.10 for each adult ticket and $7.75 for each student ticket. Mrs. Williams purchased 7 tickets
zhuklara [117]

Answer:

Two adult tickets and 5 student tickets

Step-by-step explanation:

Let a=adult tickets   Let s=student tickets

You know that each adult ticket is $9.10 and each student ticket cost $7.75. At the end, it cost $56.95 for both students and adults so the first equation should be 9.10a+7.75s=56.95. To get the second equation, you know that Mrs. Williams purchased 7 tickets in total that were both students and adults. Therefore, the second equation should be a+s=7. The two equations are 9.10a+7.75s=56.95

a+s=7.

Now, use substitution to solve this. I will isolate s from this equation so the new equation should be s=-a+7. Plug in this equation to the other equation, it will look like this 9.10a+7.75(-a+7)=56.95. Simplify this to get 9.10a-7.75a+54.25=56.95. Simplify this again and the equation will become 1.35a=2.70. Then divide 1.35 by each side to get a=2. This Mrs. Williams bought two adult tickets. Plug in 2 into a+s=7, it will look like this (2)+s=7. Simplify this and get s=5. This means Mrs. Williams bought five adult tickets. Therefore she bought 2 adult tickets and 5 student tickets.

Hope this helps

4 0
3 years ago
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