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3 years ago

Simplify ^3_/125x^21y^33

1 answer:

7 0

$\sqrt[3]{125x^{21}y^{33}}=\sqrt[3]{125}\cdot\sqrt[3]{x^{21}}\cdot\sqrt[3]{y^{33}}=\sqrt[3]{5^3}\cdot\sqrt[3]{(x^7)^3}\cdot\sqrt[3]{(y^{11})^3}\\\\=\boxed{5x^7y^{11}}\\\\Used:\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt[n]{a^n}=a\\\\(a^n)^m=a^{nm}$

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Elena received a $30 gift for her birthday. She plans to use it to buy one or more books. Elena will have to pay 8% sales tax on

11111nata11111 [884]

The cost of book has to be less than or equal to 26.39.

The amount of money Elena can buy has to be less than or equal to $30. Thus, the inequality sign to be used is the less than or equal to sign.

(1 + tax)cost of books + cost of bus fare ≤ value of the gift

1.08c + 1.5 ≤ 30

In order to determine the maximum cost of books, take the following steps:

Combine similar terms

1.08c ≤ 30 - 1.5

Add similar terms together

1.08c ≤ 28.50

Divide both sides of the equation by 1.08

c ≤ 28.50 / 1.08

c ≤ 26.39

To learn more about taxes, please check: brainly.com/question/25311567

3 0

2 years ago

If 55% people in a city like Cricket, 30% like Football and the remaining like nother games, then what per cent of the people li

labwork [276]

$\large{\dag\:{\underline{\underline{\frak{\pmb{\red{Answer:-}}}}}}}$

Cricket = 55%
Football = 30%
Other games = Remaining (?)

So, the percent of people who like other games equals:

= 100 – (55 + 30)

= 100 – 85

= 15%

If the total no. of. people is 60,00,000:

★ Cricket

= 60,00,000 × 55/100

= 33,00,000

★ Football

= 60,00,000 × 30/100

= 18,00,000

★ Other sports

= 60,00,000 × 15/100

= 9,00,000

6 0

2 years ago

10.2 = 0.6y what is y

AVprozaik [17]

Answer:

y = 17

Step-by-step explanation:

isolate the variable by dividing each side by factors that don't contain the variable.

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3 years ago

A sum of Rs.750 is in the form of denominations of Rs.50 and Rs.100. If the number of 100-rupee notes is twice the number of 50-

Naddik [55]

Answer:

100 rupee notes= 6, 50 rupee notes=3

Step-by-step explanation:

3 x 50 = 150

6 x 100 = 600

600 + 150 = 750

and 3 x 2 = 6, so the amount of 100 rupee notes are twice the amount of 50 rupee notes.

Hope this helps :)

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3 years ago

What is the problem of this solving?!

nika2105 [10]

This question can be solved primarily by L'Hospital Rule and the Product Rule.

$y= \lim_{x \to 0} \frac{x^2cos(x)-sin^2(x)}{x^4}$

I) Product Rule and L'Hospital Rule:

$y= \lim_{x \to 0} \frac{[2xcos(x)-x^2sin(x)]-2sin(x)cos(x)}{4x^3}$

II) Product Rule and L'Hospital Rule:

$y= \lim_{x \to 0} \frac{[-2xsin(x)+2cos(x)]-[2xsin(x)+x^2cos(x)]-[2cos^2(x)-2sin^2(x)]}{12x^2} \\ y= \lim_{x \to 0} \frac{2cos(x)-4xsin(x)-x^2cos(x)-2cos^2(x)+2sin^2(x)}{12x^2}$

III) Product Rule and L'Hospital Rule:

$]y= \alpha + \beta \\ \\ \alpha =\lim_{x \to 0} \frac{-2sin(x)-[4sin(x)+4xcos(x)]-[2xcos(x)-x^2sin(x)]}{24x} \\ \beta = \lim_{x \to 0} \frac{4sin(x)cos(x)+4sin(x)cos(x)}{24x} \\ \\ y = \lim_{x \to 0} \frac{-6sin(x)-4xcos(x)-2xcos(x)+x^2sin(x)+8sin(x)cos(x)}{24x}$

IV) Product Rule and L'Hospital Rule:

$y = \phi + \varphi \\ \\ \phi = \lim_{x \to 0} \frac{-6cos(x)-[-4xsin(x)+4cos(x)]-[2cos(x)-2xsin(x)]}{24x} \\ \varphi = \lim_{x \to 0} \frac{[2xsin(x)+x^2cos(x)]+[8cos^2(x)-8sin(x)]}{24x}$

V) Using the Definition of Limit:

$y= \frac{-6*1-4*1-2*1+8*1^2}{24} \\ y= \frac{-4}{24} \\ \boxed {y= \frac{-1}{6} }$

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3 years ago