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2 years ago

The degree of polynomial is ?

1 answer:

6 0

The degree of an individual term of a polynomial is the exponent of its variable; the exponents of the terms of this polynomial are, in order, 5, 4, 2, and 7. The degree of the polynomial is the highest degree of any of the terms; in this case, it is 7.

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Making coconut cookies. The recipe calls for 420 g of coconut and 120 g of sugar. Only has 252 g of coconut. How much sugar must

geniusboy [140]

Answer: 72 grams.

Step-by-step explanation:

Let be "s" the amount of sugar in grams that must be used for 252 grams of coconut to keep the same ratio of coconut to sugar.

Knowing that in the recipe for the coconut cookies, should be 420 grams of coconut and 120 grams of sugar, and you only have 252 grams of coconut, you can set up this proportion to find "s":

$\frac{420grams}{120grams}=\frac{252grams}{s}$

Now, you need to solve for "s":

$\frac{420grams}{120grams}=\frac{252grams}{s}\\\\s(\frac{420grams}{120grams})=252grams\\\\s=(252grams)(\frac{120grams}{420grams})\\\\s=72grams$

4 0

3 years ago

What is the probability of rolling a 4 on a single sided die and then rolling a 6?

Rzqust [24]

Answer:

If it is a six-sided die then it has the probability of 1/6 or .16

Step-by-step explanation:

4 0

3 years ago

Hannah has a aquarium with 25 Liter of water in it.How many milliliters of water are in the aquarium

Vikentia [17]

Answer:

25000

Step-by-step explanation:

25 * 1000 = 25000

8 0

2 years ago

Read 2 more answers

144x^{2}[/tex]-169<img src="https://tex.z-dn.net/?f=-y%5E%7B2%7D" id="TexFormula1" title="-y^{2}" alt="-y^{2}" align="absmiddle"

Ainat [17]

Answer:

144x^2−y^2−169 because ther are no like terms

Step-by-step explanation:

5 0

3 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

3 years ago