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3 years ago

Find the product. (-3ab2)3

Mathematics

1 answer:

Alex17521 [72]3 years ago

7 0

Answer:

answer is -27 a^3b^6

Step-by-step explanation:

use power rules (b^n)^m = b^(n⋅m)

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What is the sum of 3/7, 4/5, and 2/3 rounded to two decimal places

bagirrra123 [75]

Hello!!

Total = 3/7 + 4/5 + 2/3

= 0.428571429 + 0.8 + 0.66666667

= 199/105

= 1.89524.....

≈ 1.90 (rounded to two decimal places)

Good luck :)

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3 years ago

Jack's Plumbing charges customers $100 per hour for repair jobs. A-1 Plumbing charges according to

s2008m [1.1K]

Answer:

37/3

Step-by-step explanation:

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3 years ago

I u asnwer this u are gay

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Answer:

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3 years ago

Read 2 more answers

The following is the converse of a conditional statement. If it is not April 17th, then it is not Matt's birthday. What is the i

user100 [1]

<span>The original statement would remove the two "not" values from the converse, giving "If it is April 17th, then it is Matt's birthday." Similarly, to find the inverse of the original conditional, simply flip the two statements. As such, the inverse would read "If it is Matt's birthday, then it is April 17th."</span>

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4 years ago

Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1

aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

${3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}$

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

$3y^{2}.\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x}{y^{5}}$

3 0

3 years ago