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3 years ago

What is the square root of 9801??

Mathematics

2 answers:

professor190 [17]3 years ago

4 0

<span>the square root of 9801 is 99</span>

Ivan3 years ago

4 0

The square root of 9801 would be 99.

hope that helped

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(4x-1) /2=x+7<br>4x-1 all over 2 = x +7

Damm [24]

Answer:

x = $\frac{15}{2}$

Step-by-step explanation:

Given

$\frac{4x-1}{2}$ = x + 7 ( multiply both sides by 2 to clear the fraction )

4x - 1 = 2(x + 7) ← distribute

4x - 1 = 2x + 14 ( subtract 2x from both sides )

2x - 1 = 14 ( add 1 to both sides )

2x = 15 ( divide both sides by 2 )

x = $\frac{15}{2}$

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3 years ago

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Jia is skipping a rock across a pond. The sequence {4.2, 3.57, 3.0345, 2.5793, …} describes the height of the rock on each succe

mylen [45]

Answer:

$u_n=4.2(0.85)^{n-1}$

Step-by-step explanation:

$u_1=4.2\\u_2=3.57\\u_3=3.0345\\u_4=2.5793$

Geometric formula sequence: $u_n=ar^{(n-1)}$

(where $a$ is the first term of the sequence and $r$ is the common ratio)

To find the common ratio, divide one of the terms by the previous term:

$r=\frac{u_2}{u_1} =\frac{3.57}{4.2} =0.85$

From inspection, $a=4.2$

Therefore, $u_n=4.2(0.85)^{n-1}$

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2 years ago

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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:

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

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

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3 years ago