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OleMash [197]
3 years ago
12

5 Carla packs 4 pairs of pants, 2 pairs of shorts, and 3 skirts for her vacation.

Mathematics
2 answers:
ratelena [41]3 years ago
8 0

Answer:

D 3:4

Step-by-step explanation:

she packs 3 skirts and 4 pants so it's 3:4. The wording of the question is important (skirts to pants) so you have to put the number of skirts first before pants which is why it IS NOT A.

Murljashka [212]3 years ago
3 0
The correct answer is D
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A number t is less than or equal to 5
Pavel [41]
It would be written as
T < 5, but there would be a line underneath the less then sign
4 0
3 years ago
Which of the following expressions represents the statement below?
exis [7]

Option D. 2(n + 9) is the right answer

Step-by-step explanation:

The word problems are converted into mathematical problems to solve simply.

for the given problem

"two times the sum of a number and 9"

Let n be a number then

when there is and between two things they are performed first then next operation is performed so

the sum of a number and 9

n+9

Then

two times the sum of a number and 9

2(n+9)

Hence,

Option D. 2(n + 9) is the right answer

Keywords: Mathematical notation

Learn more about mathematical notation at:

  • brainly.com/question/10364988
  • brainly.com/question/10435816

#LearnwithBrainly

5 0
3 years ago
Rickey has 7 over 8 cup of cereal in a bowl. He then gave two 1 over 4 cups of cereal from his bowl to his little sister. Which
NeX [460]

Answer:

5/8 cup left

Step-by-step explanation:

7/8 - 1/4

1/4=2/8

7/8-2/8

5/8

5 0
3 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
Round 4.9451 to the nearest thousandths
maria [59]

Answer:

49.451 this would be the answer

5 0
2 years ago
Read 2 more answers
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