civonamzuk

Just as Jugoslav pointed out, there's a mistake in this question; namely, instead of 18 cm, should be 180 cm!

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

The total length of the lines in the 1st square is 180 cm.

The total length of the lines in the second square is 72 cm.

Luis and Jack had 350 marbles altogether. Luis gave away 25% as many marbles as Jack. Jack was left with 40 more than the number of marbles he gave away? Luis had 25% as many marbles as Jack in the end. How many marbles did Jack have in the end?

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

L + J = 350

J gave away = X

L gave away = 1/4(X)

J had in the end = X + 40

L had ............... = 1/4(X + 40)

350 - (X + X/4) = (X + 40) + 1/4(X + 40)

X = 120

X + 40 = 160

Jack had 160 marbles

Angle X = 180 - (120 + 45) = 15

XY / XZ = sin(Z) / sin(Y)

XY = (4 * sin120) / sin45

XY ≈ 4.9 

Let ABCD be a trapezoid with bases AB and CD.  Let AD = 5 and BC = 7, and let P be a point on side CD such that CP/PD = 7/5.  If Q is a point on AB such that PQ is parallel to AD, then find AQ/QB.

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

https://web2.0calc.com/questions/trapezoid_11

Find AC such that AB= 5sqrt2 and angle b= 45 and angle a=15.

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

AD = (5√2) * sin(45º)

AC = AD / cos(30º)

N : H = 1 : 3

(x + 4) / (3x + 3) = 1/2

x = 5

Hannah had 3*5+3 = 18 apples 

Angle ACP = 12 degrees

Not a problem. 

Check your answer again. Why didn't you use the quadratic formula to solve this problem?!?

36 - 4.144 = 31.856

4.144 * 31.856 = 132.011264

Let line AB and line CD be chords of a circle, that meet at the point Q inside the circle. If AQ = 16, BQ = 12, and CD = 36, then find the minimum length of CQ. 

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

I didn't see the word integer

AQ * BQ = CQ * DQ

16 * 12 = CQ(36 - CQ)

CQ = 2(9 - √33)