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What does x equal in...

\(-4x-8=-12x+6\)     on both sides: +12x + 8

           \(8x=14\)                  on both sides: divide by 8

             \(x=1\frac{3}{4}\)

  !

Cphil can u help me?

Shouldn't it just be \(\boxed{25\pi}\)

3) In triangle ABC, AB=17, AC=*, and BC=15. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD.

If AC was supposed to be 8, we have a right triangle....

Its area  = (1/2) product of the legs  = (1/2) AC * BC  =  (1/2) 8 * 15  =  60  units^2

2) In a certain iscoceles right triangle, the altitude to the hypotenuse has length 4sqrt2. What is the area of the triangle?

Area  =  (1/2)hypotenuse * (altidude drawn to the hypotenuse)   (1)

Area  (1/2)  product of the leg lengths    (2)

Equate (1) and (2)  and let the leg lengths  = L  and we have

(1/2) hypotenuse * ( 4sqrt 2 )  = (1/2) L^2     multiply through by 1/2

hypotenuse * ( 4sqrt(2)   =  L^2

The hypotenuse  =  sqrt  ( L^2 + L^2)  =  sqrt (2 L^2 )  =   Lsqrt 2

L sqrt 2 * 4* sqrt 2  = L^2    divide through  by L

4 sqrt 2 * sqrt 2  = L

4*2  = L

8 = L

So...the area  =  (1/2) L^2  = (1/2) * 8^2  =  32 units^2

\(ab^4 = 12,~a^5b^5=7776 \\ \text{we want to get rid of }b \text{ by dividing a power of the first equation by a power of the 2nd}\\ \dfrac{(ab^4)^5}{(a^5b^5)^4}=\dfrac{(12)^5}{(7776)^4} \\ \dfrac{a^5b^{20}}{a^{20}b^{20}} = \dfrac{1}{a^{15}} = \dfrac{(2^23)^5}{(2^53^5)^4}=\dfrac{2^{10}3^5}{2^{20}3^{20}}=\dfrac{1}{2^{10}3^{15}}\\ a = (2^{10}3^{15})^{1/15} = 2^{2/3}3\)

Solving by substitution means that you solve for one variable in terms of the other from one equation, 

and then substitute that into the 2nd equation to actually solve for values.

\(y = -18x - 11 \\ \text{Now we substitute this into the 2nd equation}\\ -18x-11 = -22x+15 \\ \text{Now we have an equation in one variable and we solve it as usual}\\ 4x=26 \\ x=\dfrac{26}{4} = \dfrac{13}{2}\\ y=-18\left(\dfrac {13}{2}\right) - 11 = -128\)

Thank you Hectictar

Thank you!

2325

idk if this is correct

x^2 / 9  +  y^2 /25  = 1    can be transformed to  25x^2 + 9y^2  = 225    (1)

y  = 4x + k    (2)

The slope of a tangent line at any point on (1)  can be found as

50x + 18y y'  = 0

y'  =  -50x / [ 18y]   =  -25x / [ 9y]

And we are looking for  where the slope of a tangent line  = 4

So

-25x / 9y  = 4

-25x = 36y

y = (-25)/(36) x       sub this into  (1) for y

25x^2  + 9 (-25/36 x)^2 =  225

25x^2  + 9 (625/1296)x^2  = 225

4225/144 x^2  = 225 

x^2  =225 * 144 / 4225     take both roots

x = 15 * 12 / 65  = 180/65  =36/13

Or

x = -36/13

Subbing either value into  (1) to find  y we have

25 (36/13)^2  + 9y^2  = 225

32400 / 169  + 9y^2  = 225

32400 / 169 + 9y^2  = 38025/169

y^2  = [38025 - 32400 ] / [ 9 * 169]

y^2 = [5625] / [ 9 * 169]      take both roots

y = 75 / [ 3 * 13 ] =  75 / 39  = 25 / 13

OR

y = -25/13

So....the slope of the tangent line to the ellipse = 4  at   (-36/13 , 25/13)  and (36/13. -25/13)

Writing an equation of one tangent line using the first point we have

y  = 4 ( x + 36/13)  + 25/13

y = 4x + 144/13 + 25/13

y = 4x   + 169/13

y = 4x + 13

And writing the equation of the other tangent line we have that

y = 4 (x - 36/13) - 25/13

y = 4x - 144/13  - 25/13

y = 4x -169/13

y = 4x - 13

Note the graph here :  https://www.desmos.com/calculator/syelncmges

When k  = 0 .....the graph intersects the ellipse at two points

However when k < -13  ...the tangent line is shifted to the right of the ellipse

And when k > 13....the tangent line is shifted to the left of the ellipse

See the answer here:  https://web2.0calc.com/questions/help-please_77693

No prob  !!!

you are correct, thank you very much

I'll give this a shot

We can transform the given equation  into

9x^2  + 4y^2   = 36

This is an ellipse centered at the origin

We have

9x^2  - 18x + 4y^2 + 16y  =  c    ..... [ complete the square on x , y]

9 ( x^2 - 2x + 1)  + 4 ( y^2 + 4y + 4)  = c + 9 + 16

9 ( x - 1)^2  +  4 (y + 2)^2  =   [ c + 9 + 16 ]     divide both sides by 36

(x - 1)^2 /  4   +  (y + 2)^2 / 9  =  [ c + 9 + 16] / 36

Since we want the right side to be  1,  c  must  = 11

So  we have the equation

(x - 1)^2 / 4  +  (y + 2)^2 / 9   =  1

So....the translation of the original graph  is  1 unit to the right and  2 units down  =   vector ( 1 , -2 )

So...a =  1 , b = -2   and  c  = 11

yeah so sorry, i managed to write the question wrong

That isn't the equation of an ellipse...it's the equation of a hyperbola....

Thank you!

Rom would have answered you long ago if you had proofed your question properly in the first place.

1. Given f(x)=3ax^2+2bx+x with a≠0, when 0≤x≤1 |f(x)|≤1, please determine the maximum value of a. 

f(0)=0

f(1)=3a+2b+1

\(-1\le3a+2b+1\qquad and \qquad 3a+2b+1\le+1\\ -2\le3a+2b\qquad and \qquad 3a+2b\le0\\ -2-2b\le3a\qquad and \qquad 3a\le-2b\\ \frac{-2-2b}{3}\le a\qquad and \qquad \qquad a\le\frac{-2b}{3}\\ \frac{-2b}{3}-\frac{2}{3}\le a\qquad and \qquad \qquad a\le\frac{-2b}{3}\\ \frac{-2b}{3}-\frac{2}{3}\le a\le\frac{-2b}{3}\\ \)

I want the biggest 'a' so it follows that  

\(a=\frac{-2b}{3}\\ b=-\frac{3a}{2}\)

It also follows that a is positive and b is negative

so the question becomes

\( f(x)=3ax^2+2bx+x \\  f(x)=3ax^2+2*\frac{-3a}{2}x+x \\  f(x)=3ax^2-3ax+x \\ or\\  f(x)=x(3ax-3a+1) \)

with a≠0, when 0≤x≤1 |f(x)|≤1, please determine the maximum value of a. 

and f(1)= 3a+2b+1 = 3a-3a+1 = 1

So f(x) is a concave up parabola passing through (0,0) and (1,1) 

\(f'(x)=6ax-3a+1\\ stat\;pt \;is\;when\; f'(x)=0\\ 6ax-3a+1=0\\ 6ax=3a-1\\ x=\frac{3a-1}{6a}\\ or\\ x=\frac{1}{2}-\frac{1}{6a}\\ \)

Now a is positive and as big as possible so I am going to assume that an 'a' exists such that

the stationary point will lie between    x=0 and x=0.5  

This means that the stationary point (minimim) lies in the designated region of the graph.

So the minimum value must be greater than or equal to -1   (we already know it is less than 1)

The minimum value of f(x) is

\(f(\frac{3a-1}{6a})=\frac{3a-1}{6a}[(3a*\frac{3a-1}{6a})-3a+1]\ge-1\\ \frac{3a-1}{6a}[(\frac{3a-1-6a+2}{2})]\ge-1\\ remembering \;that\;a>0\\ (3a-1)(3a-1-6a+2)\ge-12a\\ (3a-1)(-3a+1)\ge-12a\\ (3a-1)(3a-1)\le12a\\ 9a^2-6a+1\le12a\\ 9a^2-18a+1\le0\\ \text{If this was set to y it would be a concave up parabola so} \\\text{ the function will be less than zero between the roots. } \\\text{Find the roots.}\\ a=\frac{18\pm\sqrt{324-36}}{18}\\ a=1\pm\frac{2\sqrt{2}}{3}\\ \text{So the biggest value of a is}\\ a=1+\frac{2\sqrt{2}}{3}\\ a\approx 1.9428 \)

Here is the graph

Yes exactly from definition of  invertible matrix!

Thanks guys for your tips! 

Yes I think you got it, But just to be sure:

You got the equation A*(-A³+A²-A¹+I)=I. We can replace -A³+A²-A¹+I with "B". So we got A*B=I. I is an invertible matrix, and if a multiplication of two matrices results in an invertible matrix this means that both matrices are also invertible, meaning that A is invertible.

Its -A^4 + A^3 - A^2 + A = I <=> 

<=> A( -A^3 + A^4 - A + I) =I 

so ( -A^3 + A^4 - A + I)=A^-1