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4 years ago

Please help NOBODY WELL HELP

Mathematics

2 answers:

7nadin3 [17]4 years ago

4 0

The answer to this question is A

goblinko [34]4 years ago

3 0

Your equation is:
y = 7x + 126.
X represents each extra box the costumer buys.

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7-41= (-15)+33= 62-84= (-26)-14=

Gala2k [10]

Answer:

-34

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Explanation:

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4 years ago

Alice Correa bought three yards of cloth to make a dress. The cloth was on sale for $1.93 per yard. How much did Alice pay for t

vitfil [10]

The correct answer for this question is that Alice Correa will have spent $6.07 on the cloth overall. This can be worked out through first considering what 5% of the original price will be, so $5.79 / 100 = 0.0579 x 5 = $0.28, so $5.79 + $0.28 = $6.07

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3 years ago

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Which of the following will be constructed when the two endpoints of a line segment are folded so that they line up

nataly862011 [7]

B. Midpoint of a line segment

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4 years ago

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Simplify 6 to the 3rd

elixir [45]

Answer:

Isnt it 2?

Step-by-step explanation:

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3 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