All Answers 355923 Answers

338,144 books have been sold to date.

Let the total sales = S, then:

9.7%S = 32,800

0.097S= 32,800        divide both sides by 0.097

I will leave it to you to get the final answer.

Thanks so much for the responses Heureka!

(b) The circle 𝑥2 + 𝑦2 − 4𝑥 − 6𝑦 + 𝑘 = 0 touches the x-axis. Find k and the
coordinates of the point of contact. Find the coordinates of the centre and the radius of the circle.

\(k=\ ?\)

\(\begin{array}{|rcll|} \hline x^2+y^2-4x-6y +k &=& 0 \quad & | \quad y = 0 (\text{ circle touches the x-axis} ) \\ x^2 -4x +k &=& 0 \\ x &=& \frac{4\pm \sqrt{16-4k} }{2} \quad & | \quad \text{ only one solution for x} \\ 16-4k &=& 0 \\ 4k &=& 16 \\ k &=& 4 \\\\ x &=& \frac{4\pm 0 }{2}\\ x &=& 2 \\ \hline \end{array} \)

\( k = 4\\\text{The coordinates of the point of contact is $ (2,0)$}\)

\(\text{Coordinates of the centre $(c_x, cr_y)$ and the radius $r$ of the circle.} \)
\(\begin{array}{|rcll|} \hline x^2+y^2-4x-6y +k &=& 0 \\ x^2-4x+y^2-6y +k &=& 0 \\ (x-2)^2-4+(y-3)^2 -9 +k &=& 0 \\ (x-2)^2 +(y-3)^2 &=& 4+9-k \quad & | \quad k = 4 \\ (x-2)^2 +(y-3)^2 &=& 4+9-4 \\ (x-2)^2 +(y-3)^2 &=& 9 \\ (x-\underbrace{2}_{=\text{c_x}})^2+(y-\underbrace{3}_{=\text{c_y}})^2 &=& \underbrace{9}_{=r^2} \\\\ c_x &=& 2 \\ c_y &=& 3 \\ r &=& \sqrt{9} = 3 \\ \hline \end{array}\)

\(\text{The coordinates of the centre $(2,3)$} \\ \text{Radius of the circle $r = 3$ } \)

1. 

(a) The line 𝑥 + 2𝑦 = 0 meets the circle x² + y² − 4𝑥 + 2𝑦 = 0 at A and B. Find the length of AB.

\(\begin{array}{|rclrcl|} \hline x+2y &=& 0 \\ x &=& -2y \\\\ x^2 + y^2 - 4x + 2y &=& 0 \quad & | \quad x=-2y \\ (-2y)^2 + y^2 - 4(-2y) + 2y &=& 0 \\ 4y^2 + y^2 + 8y + 2y &=& 0 \\ 5y^2 + 10y &=& 0 \\ 5y(y+2) &=& 0 \\\\ 5y &=& 0 & y+2 &=& 0 \\ y &=& 0 & y &=& -2 \\\\ x &=& -2\cdot 0 & x&=& -2(-2) \\ x &=& 0 & x&=& 4 \\\\ & & A(0,0) &&& B(4,-2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline AB &=& \sqrt{(0-4)^2+(0-(-2))^2} \\ &=& \sqrt{16+(0+2)^2} \\ &=& \sqrt{16+4} \\ &=& \sqrt{20} \\ &=& \sqrt{4\cdot 5} \\ &=& 2\sqrt{ 5} \\ \hline \end{array}\)

Yes the problem is with the askers not the answerers.

----

 That’s bullshitt!!   It is not just the askers. Several times you’ve reamed out the brats on here who have posted multiple questions.  Your post is directed toward the answerers because the answerers enable this behavior, and there is one in particular who does this. 

The reason there are so many uninterested homework moochers on here is that someone started feeding the bears. There is a reason for not feeding the bears. Once you feed a bear, it will return for more. The only way to get rid of it is to quit feeding it. If you keep feeding it, other bears join in the free food fest. They will eat everything you give them and one day they will eat you. 

Beggars and panhandlers are the same kind of animal. Some might say please and thank you, but they don’t really mean it. They are just magic words learned as little children. Some could actually learn how to fish, but why should they, when they see their brothers and sisters get all the fish they want just by begging? “Pander to me, I’m entitled,” is their mantra.

If this irritates you, you could hide their questions or lock them. You’ve done that a few times, but it doesn’t stop the one person mainly responsible for all the bear shitt on the forum’s sidewalks, because he’s a moderator too, and just unhides or unlocks them.

CPhill’s obsession with answering the panhandler’s questions, and completing ones that are half finished, may irritate the shitt out of you, but I like reading his answers, and the answers of other competent mathematicians.

I don’t give a flyingdamn if the dumbshit kids learn anything or not. The option is there for them if they want to learn. if not, who should care?  A lot of them are only a step or two from the retardos on here who obsessively make help posts. Trying to teach them math is a waste of time.  But others can learn from these posts, even if the asker does not, so it’s not a total waste.

I have nothing better to do than tend to my garden, drink some booze, pissa few people off on Reddit, come here, pissa few people off and learn some math. This is a great place. It’s better than Reddit, anyway. They are so crude and vulgar, I feel like I need to take a bath in bleach after I log off.   

To show my gratitude, I’ll tell you how you can reduce the bearshit on here.  You are a moderator, so just send the posts and the bearshit “down the drain.......into the sewer”  I’m sure you know what that means, the other moderators know too, unless they are idiots, and I doubt it.  Well Asinus is, but whatever! CPhill can wake sleeping beauty, but he can’t raise the dead. If Sisyphus doesn’t have his boulder, he can’t push it up the hill. ROFLMFAO!

Don’t worry, CPhill, Melody doesn’t have the cojones or the stomach for it, anyway.

\(200\pi=628.3185307179586477\)

The immediate integer that comes before this value is 623, and that's precisely how many positive integers there are below ..

\(200\pi\)

1:  Let triangle PQR be a triangle in the plane, and let S be a point outside the plane of triangle PQR, so that SPQR is a pyramid whose faces are all triangles. Suppose that every edge of SPQR has length 18 or 41, but no face of SPQR is equilateral. Then what is the surface area of SPQR?

Heron:

\(\begin{array}{|rcll|} \hline A &=& 4\times \sqrt{s(s-41)(s-41)(s-18)} \quad & | \quad s=\frac{41+41+18}{2}=50 \\ &=& 4\times \sqrt{50\cdot(50-41)^2\cdot (50-18)} \\ &=& 4\times \sqrt{50\cdot 9^2\cdot 32} \\ &=& 4\cdot9 \sqrt{50\cdot 32} \\ &=& 36\sqrt{1600} \\ &=& 36\cdot 40 \\ &=& 1440 \\ \hline \end{array}\)

2:  A rectangular box is 8 cm thick, and its square bases measure 32 cm by 32 cm. What is the distance, in centimeters, from the center point P of one square base to corner Q of the opposite base? Express your answer in simplest terms.

\(\begin{array}{|rcll|} \hline PQ &=& \sqrt{(\frac{32}{2})^2+ (\frac{32}{2})^2 + 8^2 } \\ &=& \sqrt{16^2+ 16^2 + 8^2 } \\ &=& \sqrt{576 } \\ &=& 24\ cm \\ \hline \end{array}\)

Dave's sister baked  dozen pies of which a third contained chocolate, a quarter contained marshmallows, a sixth contained cayenne, and one twelfth contained salted soy nuts. What is the smallest possible number of pies that had none of these ingredients?

3:  Let ABCDEFGH be a cube of side length 5, as shown. Let P and Q be points on line AB and line AE, respectively, such that AP = 2 and AQ = 1. The plane through C, P, and Q intersects line DH at R. Find DR.

1)A cylinder is sliced such that the cross section is perpendicular to his base.

What ist ehe shape of the cross section           ?

rectangle

2)Wich three-dimensional figure is formed by the rotation given?

Number 4

3)A company is creating a box without a top from a piece of cardboard, but cutting out square corners with side length x.

Wich expression can be used to determin the gratest possible volumen of the cardoard box?

(7-2x)(11-2x)x

Did I do it right?

You mean the vector cross product!

\(\begin{array}{|rcll|} \hline A &=& \sqrt{2^2+(-10)^2+(-7)^2} \\ A &=& \sqrt{4+100+49} \\ \mathbf{A} & \mathbf{=} & \mathbf{\sqrt{153}} \\ \hline \end{array}\)

Suppose $a$, $b,$ and $c$ are positive numbers satisfying: \begin{align*} a^2/b &= 1, \\ b^2/c &= 2, \text{ and}\\ c^2/a &= 3. \end{align*} Find $a$.

\(\begin{array}{|rcll|} \hline (1) & a^2 &=& b \\ & a &=& \sqrt{b} \\\\ (2) & b^2 &=& 2c \\ & b &=& \sqrt{2c} \\\\ (3) & c^2 &=& 3a \quad & | \quad a = \sqrt{b}\\ & c^2 &=& 3 \sqrt{b} \quad & | \quad b = \sqrt{2c}\\ & c^2 &=& 3 \sqrt{\sqrt{2c}} \\ & c^2 &=& 3 \sqrt[4]{ 2c } \\ & (c^2)^4 &=& 3^4 2c \\ & c^8 &=& 162c \\ & c^7 &=& 162 \\ & c &=& \sqrt[7]{ 162 } \\ & \mathbf{c} & \mathbf{=} & \mathbf{2.06844987991} \\\\ & a &=& \frac{c^2}{3} \\ & a &=& \frac{2.06844987991^2}{3} \\ & a &=& \frac{4.27848490568}{3} \\ & \mathbf{a} & \mathbf{=} & \mathbf{1.42616163523} \\\\ & b &=& a^2 \\ & b &=& 1.42616163523^2 \\ & \mathbf{b} & \mathbf{=} & \mathbf{2.03393700979} \\ \hline \end{array}\)

To write this out algebraically, we can assign variables to the two numbers. 

First number = x, Second number = y

The product of the two numbers is x*y or just xy

The sum of the two numbers is (x+y) 

The sum of the product and the sum of two positive integers can be expressed as: 

xy + x + y, and this is equal to 454, xy + x + y = 454 

We can solve for all values of x and y, we get: 

xy + x + y = 454

We add 1 to both sides so we can factor: 

xy + x + y + 1 = 455

Then we can factor:

x(y+1)+y+1 = 455

(y+1)(x+1) = 455 = 5 * 7 * 13

Their product is the largest only when x = 90 and y = 4 

(90*4)(90+4)=33840

Other pairs is not as big, for example, when x = 12 and y = 34

(34*12)(34+12)=18768

To answer your question, the largest possible value is 33840, 

I hope this helped,

Gavin

Yes the problem is with the askers not the answerers. I know the answerers are all very generous people who just want to help facilitate learning. All I have suggested is that you think first, answer second :)

Hi Melody,

I actually did not see the new post by CPhil, but thank you for bringing this into my knowledge. Thank you to CPhil, but I actually did do these questions myself. I am sorry if I have offended anyone and thank you again to all of those who really do help me.

-Ashra

THANK YOU SO MUCH!

Good to see you back Geno!

Oh just noticed that. Yikes!