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By definition, $r = \frac{d}{2}.$

That is a very hard question!

https://web2.0calc.com/questions/pls-help-due-tmrow_7

This is a link that will help you solve the question.

I will post an answer here based on the answer in the link.

Because of the formulas given in the link, we can figure out:

225/64

= 3.515625

The radius of circle C is 6*sqrt(2).

k=5s and s+18=k-18or k=s+36 so now's+36=5s or 4s=36 or s=9,so Sam had 9,Kabir had 5*9=45 and total 9+45=54 candy .

*54*

5x = Kabir's candy

5x - 18 = 3x

5x - 18 = 3x

2x - 18 = 0

2x = 18

x = 9

So:

6x = Total

6 * (9) = Total

56 = Total

The total amount of candy that both boys have is 56 pieces of candy.

There is already an answer on the link you posted, but I will try to solve it here.

"On the day after Halloween, Kabir had 5 times as much candy as Sam. If Kabir gives 18 pieces of candy to Sam, they will each have the same amount. How many pieces of candy do they have in all?"

If we make a variable for the number of candies Sam has, s, we can put everything in terms of s when we solve this. Because we know that Kabir had 5 times as much candy as Sam, we can make that expression:

5x = Kabir's candy

Now, we can figure out the second equation needed to figure out the answer. If Kabir gives Sam 18 candies, they will have the same amount. So:

5x - 18 = 3x

(I got 3x by dividing the total amount of candy by two, because they will have the same amount of candy once the exchange is done)

Now, we can solve the equation for x.

5x - 18 = 3x

(Subtract 3x from both sides)

2x - 18 = 0

(Add 18 to both sides)

2x = 18

(Divide both sides by 2)

x = 9

Now we can figure out the total amount by multipling x by 6. As I said above, the total amount of candy is 6x.

So:

6x = Total

6 * (9) = Total

56 = Total

The total amount of candy that both boys have is 56 pieces of candy.

The minimum area is 25.

x = (3a - b)/(a + b).

2/9R = 2(R-40)

2/9R = 2R - 80

80 = 16/9R

R = 45

yayyy :))

=^._.^=

C = 2/9R

C = 2(R-40)

2/9R = 2(R-40)

Take it from there. :))

=^._.^=

Thank you. 

I tried it, and it looks amazing. :))

=^._.^=

pi/4 = .25 pi

13 pi + .25 pi = 13.25 pi     ( or  13 1/4  pi)

8l + 10 d = 28

3l + 10 d = 23       subtract this equation from the first one to get

5 l  = 5

l = laundry = 1 dollar       then use this value ineither equation to calculate d

    8 (1) + 10 d = 28

         d = dishes = 2 dollars  

Supondo que esses sejam os comprimentos laterais do retângulo :

Area = a * b

        30 mm * 50 mm = 1500 mm²

Here is a link  (let me know if the link doesn't work)

Note: This is not a proper answer.

As shown in the diagram, a rectangular piece of paper ABCD was folded along CP where P is located one-third of the way from A to B. We know the total area covered by the folded paper is 213 square inches with a 75 square inch overlapping area. Find the area of triangle B'CQ in square inches.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

[APCB'QD] = 213 in²

[CQP] = 75 in²

[PRQ] = [B'CQ]

[ADRP] = [B'CP]

[B'CQ] = 1/3(213 - 150) = 21 in²

Like so:

Positive real numbers \(x\), \(y\)  satisfy the equations \(x^2 + y^2 = 1\) and
\(x^4 + y^4 = \dfrac{17}{19}\).
Find \(xy\).

\(\begin{array}{|rcll|} \hline \left( x^2 + y^2\right)^2 &=& x^4 + 2x^2y^2 + y^4 \\ \left( x^2 + y^2\right)^2 &=& x^4 + y^4 + 2x^2y^2 \quad | \quad x^2 + y^2 = 1,~ \dfrac{17}{19} \\\\ 1^2 &=& \dfrac{17}{19} + 2x^2y^2 \\\\ 2x^2y^2 &=& 1 - \dfrac{17}{19} \\\\ 2x^2y^2 &=& \dfrac{2}{19} \quad | \quad :2 \\\\ x^2y^2 &=& \dfrac{1}{19} \quad | \quad \text{sqrt both sides} \\\\ \mathbf{ xy } &=& \mathbf{ \dfrac{1}{\sqrt{19}} } \\ \hline \end{array}\)

y =  5x + 4       (1)

x + y = 94       (2)

Rearrange (2):   y = 94 - x   (3)

Substitute (3) into (1)

94 - x = 5x + 4    Collect like terms:  6x = 90,   x = 15

Put this back into (1)  y = 5*15 + 4    so  y = 79

It looks like 2) is the answer.  

I used the symbolic mathematics part of Mathcad to do this.  I guess that with a lot of tedious manipulation by hand you would be able to show the coefficients of alpha, beta and gamma are all zero.  I didn't see any nice shortcut!