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3 years ago

Help give me the formula and the answer and plz explain so I also benefit:)

Mathematics

1 answer:

Fudgin [204]3 years ago

4 0

I think 48 cubic cm..... (Tell me if I did it wrong or not)

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What is the median of the data set represented by the dot plot?

Butoxors [25]

So ur numbers are : 3,3,4,5,5,5,6,8,8

and the median (middle number) is 5 <==

because 2 dots above the 3 means u have 2 threes....and 1 dot above the 4 means there is 1 four..and so on

6 0

3 years ago

Read 2 more answers

Pls help extra points and mark brainlist

Katyanochek1 [597]

Answer:

(12)3=36

Step-by-step explanation:

i don't know.. I think this is what the question is trying to say XD

6 0

3 years ago

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A group of 5 men are meeting for lunch.

Sladkaya [172]

Tile:

<h2>See the explanation.</h2>

Step-by-step explanation:

(a)

There were total 5 men wearing coats.

5 coats can be returned to 5 men in 5! = 120 ways.

The coats can be returned to the accurate persons only in 1 way.

Hence, the probability that each man gets the correct coat is $\frac{1}{120}$ .

(b)

At the time of returning the first coat, the hostess will have 5 choices and for the second she will have 4 choices.

Hence, in $5\times4 = 20$ ways the hostess can return the 2 coats.

There is only 1 possible case that each of the coats will return to the correct owner.

Hence, the required probability is $\frac{1}{20}$ .

6 0

3 years ago

Anna gets paid $8.75/hour working as a barista at Coffee Break. Her boss pays her $9.00/hour for creating the weekly advertiseme

vladimir1956 [14]

Answer:

i believe it would be $443.75

Step-by-step explanation:

just add 8.75 to 9.00 then multiply it by 25

8.75+9.00= 17.75 x 25= 443.75

if its a different type of equation then maybe this didn't help but i hope it does

7 0

3 years ago

CALC- limits<br> please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

3 years ago