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serg [7]
2 years ago
7

How to solve 3a+9 and a²-9 i have to find h.c.f

Mathematics
1 answer:
Lesechka [4]2 years ago
4 0

Answer:

First equation = 3a + 9

= 3( a+3 )

Second equation = a²- 9

= a² - 3 ^ 2

= (a+3) (a-3)

Hence,

HCF = ( a+ 3)

Hope this helped

All the best !!

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15. If x=a Sin2t (I+Cos2t) and y=b Cos 2t (1-Cos2t) then find<br>dy/dx at =22/7*4<br>​
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By the chain rule,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}

It looks like we're given

\begin{cases}x=a\sin(2t)(1+\cos(2t))\\y=b\cos(2t)(1-\cos(2t))\end{cases}

where <em>a</em> and <em>b</em> are presumably constant.

Recall that

\cos^2t=\dfrac{1+\cos(2t)}2

\sin^2t=\dfrac{1-\cos(2t)}2

so that

\begin{cases}x=2a\sin(2t)\cos^2t\\y=2b\cos(2t)\sin^2t\end{cases}

Then we have

\dfrac{\mathrm dx}{\mathrm dt}=4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t

\dfrac{\mathrm dy}{\mathrm dt}=-4b\sin(2t)\sin^2t+4b\cos(2t)\sin t\cos t

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4b\cos(2t)\sin t\cos t-4b\sin(2t)\sin^2}{4a\cos(2t)\cos^2t-4a\sin(2t)\cos t\sin t}

\implies\boxed{\dfrac{\mathrm dy}{\mathrm dx}=\dfrac ba\tan t}

where the last reduction follows from dividing through everything by \cos(2t)\cos^2t and simplifying.

I'm not sure at which point you're supposed to evaluate the derivative (22/7*4, as in 88/7? or something else?), so I'll leave that to you.

8 0
3 years ago
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