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Radius of the circle 'x^2+y^2=16'

= \(\sqrt{16}\)

= 4

\(\because 1<4\text{ and }2<4\)

\(\therefore \text{Point (1,2) is in the circle, and any line going through it can't be the circle's tangent}\)

\(\therefore \text{Any line going through the point (1,2) will ALWAYS intersect the circle }x^2+y^2=16\\ \;\;\;\;\text{at 2 points.}\)

If 12/pi =t, Whatb isnjvds 

This is very similar to the one I did here:

http://web2.0calc.com/questions/can-someone-please-help-me-with-this-question-and-the-one-i-just-posted-i-m-very-stuck-and-need-to-finish-these-two-questions-to-complete-my-class-please

See if you can follow my example to work this one....if you get stuck, let me know

Find a set of parametric equations for the line that passes through the given points.

(3, 7/2, -3/2), (9/2, 3/2, 1)

The vector connecting these points is given  by  < 9/2 - 3, 3/2 - 7/2, 1 - (-3/2) .  =  < 3/2, -2, 5/2 >

So...using either of the given points [ I'll use the first one ].....the vector form of the line is :

< 3, 7/2, -3/2 > + t <3/2, -2, 5/2>   =  < 3 + (3/2)t, 7/2 - 2t, (-3/2) + (5/2)t >

And the parametric equations are

x = 3 + (3/2)t

y = 7/2 - 2t

z = (-3/2) + (5/2)t

Verify that (9/2, 3/2, 1)   lies on this line when t = 1

Look at the graph, here:  https://www.desmos.com/calculator/rouxa8yruw

As x approaches both  0 and 9 from the right, y approaxhes infinity

Are you SURE the sum of their ages doesn't add up to 87?  Or maybe 80?

Yes, I meant it like that Omi67. Thank you. Can we do it by factoring out a 3 in the first term instead of working with those annoying fractions?

Ignore my last comment!

On second thoughts, the minus sign should take this into account automatically!

1.  -1, 0, 1

2.  Infinite nbr. e.g -1/2, 1/2, 3/2

3. If we interpret integer as positive whole number, then 1

4. Infinite nbr: -sqrt(2), sqrt(2), sqrt(3)

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Oops! You'll need to think a little more carefully for the angle with the positive y axis.  The method above will give the angle with the negative y axis.

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Unit vector in x-direction ux = i 

A.ux = |A|*|ux|*cos(thetax)

(3i- 6j + 2k).(i) = 3        (because i.i = 1,  j.i = k.i = 0)

|A| = sqrt(3^2 + 6^2 + 2^2) → 7

|i| = 1

Hence:  3 = 7*1*cos(thetax)

cos(thetax) = 3/7

thetax  ≈ 64.6 degrees

Do a similar calculation for y and z, bearing in mind that the unit vector in the y direction is j, that in the z direction is k, and i.j = k.j = 0, j.j = 1, k.i = k.j = 0, k.k = 1

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i shall guess, but is it zero? im not in a algbra class, just a regular math class

83.333.... Million

Let x = 0.24242424.......

100x = 24.24242424......

Subtract the first from the second:

99x = 24

x= 24/99.    Or x = 8/33

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As far as I can see there is only one real number solution:

3/4 times 4 is 3.

(-3*15)+13^2+2

(-45) + 169 + 2 = 

126

5 / (z^2 - 8z + 7 )  =  1 / [4z + 4]   cross-multiply

5 [ 4z + 4]  = z^2 - 8z + 7   simplify

20z + 20  = z^2 - 8z + 7       subtract 20z + 20 from both sides

0  = z^2 - 28z - 13

Using the quadratic formula....the solutions  are    14 ± √209