hectictar

I remember doing these D:

The slope of AB must match the slope of CD.

The slope of BC must match the slope of AD.

Slope of AB =  \(\frac{-1-6}{-5+9}=-\frac{7}{4}\)

Slope of CD = \(\frac{5+2}{-1-3}=-\frac{7}{4}\)

Slope of BC = \(\frac{6-5}{-9+1}=-\frac{1}{8}\)

Slope of AD = \(\frac{-1+2}{-5-3}=-\frac{1}{8}\)

So far so good. We have just shown that this is a parallelogram at least.

In order for it to be a rhombus:

The slope of BD must be the negative reciprocal of the slope of AC.

Slope of BD = \(\frac{6+2}{-9-3}=-\frac{8}{12}=-\frac{2}{3}\)

Slope of AC = \(\frac{-1-5}{-5+1}=\frac{-6}{-4}=\frac{3}{2}\)

Everything checks out. This figure is infact a rhombus. :)

\(-14=\frac{2(h-3)}{5}\)

Multiply both sides of the equation by 5:

\(-14(5)=2(h-3) \\~\\ -70=2(h-3)\)

Divide both sides of the equation by 2:

\(\frac{-70}{2}=(h-3) \\~\\ -35=h-3\)

Add 3 to both side of the equation:

\(-35+3=h \\~\\ -32=h\)

There are 25 hockey players. There are 5 hockey players per group.

So there are 25/5 = 5 groups.

Each group gets 30 candy bars.

If all 5 groups sold all 30 candy bars, that would be a total of 5*30 = 150 candy bars.

Since it is an isocolese trapezoid, the two nonparallel sides, aka the legs, have the same length.

AB = CD

7y - 4 = 8y - 18

7y +14 = 8y

14 = y

[  I think they just put the info about BC being 4y - 6 to try to trick you. :)  ]

\(\tan(\text{ the angle }) = \frac{ (\text{ length of side opposite the angle })}{(\text{ length of side adjacent the angle })}\)

Just plug in the information that you have and solve for whatever you need.

The reverse of tan is arctan. (This can also be expressed as atan or tan^-1.)

\(( \text{ the angle }) = \arctan(\frac{ (\text{ length of side opposite the angle })}{(\text{ length of side adjacent the angle })})\)

The reverse of sin is arcsin (or asin or sin^-1).

The reverse of cos is arccos (or acos or cos^-1).

Another thing to remember is that the adjacent side can never be the hypotenuse.

Since both of the opposite sides are parallel, angle 2 is the exact same as angle 1.

mL2 = 160º

The two unnamed angles are also equal to each other for the same reason.

The horizontal line bisects the two unnamed angles.

There are 180º in every triangle.

So

 mL2 + 2(mL3) = 180º

160º + 2(mL3) = 180º

2(mL3) = 20º

mL3 = 10º

9(6+m)=-27

Divide both sides by 9:

(6 + m) = -27/9

6+m = -3

Subtract 6 from both sides:

m = -3-6

m = -9

I mean fifth column from left.

spiders, definitely

\(\frac{12}{3}\sqrt{2} \\~\\ = 4\sqrt{2} \approx 5.657\)

Or you might have meant:

\(\frac{12}{3\sqrt{2}} \\~\\ = \frac{4}{\sqrt{2}} \\~\\ = \frac{4\sqrt{2}}{2} \\~\\ = 2\sqrt{2} \approx 2.828\)