All Answers

+219

-3

Its supposed to be spring and its freezing.. This is soo unfair!

EmeraldWonderMar 7, 2019

+444

-5

No problem ; )

HiylinLinkMar 7, 2019

+1009

-5

Thankyou

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NickolasMar 7, 2019

+444

-5

Point D should be point (5, -5) and point C should be (-4,-5) remember works just like a regular rectangle counting points wise and notice how these points are ten points down or ten units this means the area is correct because 9 x 10 is = to 90

HiylinLinkMar 7, 2019

+118653

Thanks Heureka :)

MelodyMar 7, 2019

+444

-6

No problem ; )

HiylinLinkMar 7, 2019

+279

Thank you so much Link

CreatorRozenMar 7, 2019

+444

-6

For multiplying fractions we align them next to each other you now just like adding and we look if any numerator and denominator has a common whhole number factor which is called cancling but in this case there is non so we multiply the numerator by the numerator and the same thing with the denomenators

56/33 but we need to reduce this so 33 goes in to 56 1 so we get 23/33

So the answer is 1 23/33 the one is a whole number

HiylinLinkMar 7, 2019

+26387

Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy.

x^5y^2-x^3y+5xy^3=0

\(\begin{array}{|l|rcll|} \hline & \mathbf{x^5y^2-x^3y+5xy^3} & \mathbf{=}& \mathbf{0} \\\\ \hline y'=\ ? & (5x^{5-1}\cdot y^2+ x^5\cdot 2y^{2-1}\cdot y') \\ & - (3x^{3-1}\cdot y+ x^3\cdot y') \\ & + 5(1\cdot x^{1-1}\cdot y^3 +x\cdot 3y^{3-1}\cdot y' ) &=& 0 \\ \hline & (5x^4y^2+ x^5\cdot 2yy') - (3x^2y+ x^3y') + 5(y^3 +x\cdot 3y^2y' ) &=& 0 \\ & 5x^4y^2+ 2x^5yy'- 3x^2y - x^3y' + 5y^3 +15xy^2y' &=& 0 \\ & 2x^5yy'- x^3y'+15xy^2y'&=& 3x^2y-5y^3-5x^4y^2 \\ & y'x(2x^4y-x^2+15y^2)&=& y(3x^2-5y^2-5x^4y) \\ & \mathbf{y'} &\mathbf{=}& \mathbf{\dfrac{y(3x^2-5y^2-5x^4y)}{x(2x^4y-x^2+15y^2)}} \\ \hline \end{array}\)

heurekaMar 7, 2019

+444

-6

There is actually a pretty simple way to find these and it's kind of fun or at least to me it is with knowing there are tweleve inches in a fott you multiply that by three once to get a yard then you ultiply it by the number the three yards as demonstrated like this 12x 3 = 36 and 36 x 3 = 108 and then you divide 36 inches by four to find the 1/4 a yard and we get 9 and 9 + 108 = 117.

HiylinLinkMar 7, 2019

+26387

14+14-14*14/14+15+15-15*15/15+3+3-3*3/3

\(\begin{array}{|rcll|} \hline && 14+14-14*\underbrace{\dfrac{14}{14}}_{=1}+15+15-15*\underbrace{\dfrac{15}{15}}_{=1}+3+3-3*\underbrace{\dfrac{3}{3}}_{=1} \\ &=& 14+14-14*1+15+15-15*1+3+3-3*1 \\ &=& 14+\underbrace{14-14}_{=0} +15+\underbrace{15-15}_{=0} +3+\underbrace{3-3}_{=0} \\ &=& 14+0 +15+0 +3+0 \\ &=& \mathbf{32} \\ \hline \end{array}\)

heurekaMar 7, 2019

+26387

In right triangle \(XYZ\) with \(\angle YXZ = 90^\circ\), we have \(XY = 24\) and \(YZ = 25\). Find \(\tan Y\).

\(\begin{array}{|rcll|} \hline 24^2+XZ^2 &=& 25^2 \\ XZ^2 &=& 25^2-24^2 \\ &=& (24+1)^2 - 24^2 \\ &=& 24^2 + 48 + 1 - 24^2 \\ &=& 49 \\ &=& 7^2 \\ \mathbf{XZ} &\mathbf{=} & \mathbf{7} \\\\ \tan(Y) &=& \dfrac{XZ}{24} \\ \mathbf{\tan(Y)} &\mathbf{=} & \mathbf{\dfrac{7}{24}} \\ \hline \end{array}\)

heurekaMar 7, 2019

+118653

What is your question? I mean what is the problme that you have with this question.

Please be specific. What are your thoughts on it?

MelodyMar 7, 2019

+26387

Find y'' by implicit differentiation.

5x^3 + 4y^3 = 6

\(\begin{array}{|l|rcll|} \hline & \mathbf{5x^3 + 4y^3} & \mathbf{=}& \mathbf{6} \\\\ \hline y'=\ ? & 5\cdot 3x^{3-1} + 4\cdot 3y^{3-1}\cdot y' &=& 0 \\ & 15x^2+12y^2y' &=& 0 \\ & \mathbf{y'} &\mathbf{=}& \mathbf{-\dfrac{5x^2}{4y^2}} \\ \hline y''=\ ? & 15x^2+12y^2y' &=& 0 \\ & 15\cdot2 x^{2-1}+12\cdot\left( (y^2)'\cdot y' +y^2\cdot y'' \right) &=& 0 \\ & 30 x^{2-1}+12\cdot\left( 2\cdot y^{2-1}y'\cdot y' +y^2\cdot y'' \right) &=& 0 \\ & 30 x+12\cdot\left( 2\cdot y(y')^2 +y^2\cdot y'' \right) &=& 0 \quad |\quad : 6 \\ & 5x+ 2\cdot\left( 2\cdot y(y')^2 +y^2\cdot y'' \right) &=& 0 \\ & 5x+ 4y(y')^2 + 2y^2\cdot y'' &=& 0 \quad |\quad \mathbf{y'=-\dfrac{5x^2}{4y^2}}\\ & 5x+ 4y\left(-\dfrac{5x^2}{4y^2}\right)^2 + 2y^2\cdot y'' &=& 0 \\ & 5x+ \dfrac{4y\cdot 25x^4}{16y^4} + 2y^2\cdot y'' &=& 0 \\ & 5x+ \dfrac{25x^4}{4y^3} + 2y^2\cdot y'' &=& 0 \quad |\quad \cdot 4y^3 \\ & 20xy^3+ 25x^4 + 8y^5\cdot y'' &=& 0 \\ & 5x(\underbrace{4y^3+ 5x^3}_{=6}) + 8y^5\cdot y'' &=& 0 \\ & 5x\cdot 6 + 8y^5\cdot y'' &=& 0 \\ & 30x + 8y^5\cdot y'' &=& 0 \quad |\quad :2 \\ & 15x + 4y^5\cdot y'' &=& 0\\ & 4y^5\cdot y'' &=& -15x \quad |\quad :4y^5 \\ & \mathbf{y''} &\mathbf{=}& \mathbf{-\dfrac{15x}{4y^5}} \\ \hline \end{array}\)

heurekaMar 7, 2019