SparklingWater2

1.

we see that half of the original triangle is a 3-4-5 triangle, so $\frac{4\cdot\frac{13.5}3\cdot27}2=\boxed{243}$

2.

we know that two triangles are the same if they share the same intercepted arc and that one of the angles are similar, so the triangles are similar. this means $\frac46=\frac{10}{CD}$

$4CD=60$

$CD=\boxed{15}$

3.

we can draw a line MN, that intersects at point R, which is also the midpoint of AC and NM. PQ/AC=PR/AR.

we see that PRM is similar to DPA because of the angles, and we know R is the midpoint of NM which is parallel to AD, so triangle DPA's lengths are 2 times the lengths of PRM, so PR/AP=PQ/AC=$\boxed{\frac12}$

lets say she buys m killograms of M&M's, and p killograms of peanuts. we have:

5m+7.5p=72.5

m+p=10

5m+5p=50

2.5p=22.5

p=9

m+9=10

m=$\boxed{1}$

$1\cdot2+(6-2)\cdot(6+1)+\frac{(10-6-1)\cdot(6-2)}2=2+28+6=\boxed{36}$

1.

we see that half of the original triangle is a 3-4-5 triangle, so $\frac{4\cdot\frac{13.5}3\cdot27}2=\boxed{243}$

2.

we know that two triangles are the same if they share the same intercepted arc and that one of the angles are similar, so the triangles are similar. this means $\frac46=\frac{10}{CD}$

$4CD=60$

$CD=15$

(I'll do third after class)

1.

we can first start off by figuring out how many degrees are in each interior angle of a hexagon: (6-2)*180$^\circ$=720$^\circ$           $\frac{720^\circ}{6}=120^\circ$  and we know that the triangle with B and C as the two base vertices is an isoscolese, because the square and the hexagon have equal side lengths, so angle B = angle C. so we have

$\frac{180-(120-90)}/2$= angle C = $75^\circ$

and because we want to find angle ABC, we have to subtract 75$^\circ$ from 120 to get angle ABC = 120-75 = $\boxed{45^\circ}$

2.

Because of power of point, we know that PC$\cdot$CD = PA$\cdot$AB = 25 = 4AB

so AB=$\boxed{\frac{25}4}$

3.

The diagonal of rectangle ABCD is the diameter of the circle, so

4$^2$+6$^2$= CD^2 = 52

CD= 2$\sqrt{13}$

so the area of the square is:

$(\frac{2\sqrt{13}}2)^2\cdot\pi=13\pi$

so

shaded region = $\boxed{13\pi-24}$

No problem!

whoops... didn't see it said "less"... thank you CPhill

x-3=4

add 3 to both sides

x=$\boxed{7}$

If NM is the median of trapezoid ABCD, then that means NM is the average of the two heights, which in this case are 5 and 17. so that means NM= $\frac{5+17}2=11$

now to find the median of trapezoid ABMN, we have to do the same thing, take the average of the two heights, which in this case are 5 and 11. This means that PQ=$\frac{5+11}2=\boxed{8}$

690/2 is the length of the width+length, which is 345. if w=2l-30 and w+l=345, then subtracting one equation from the other, we have

l=-2l+375

3l=375

length=$\boxed{125}$ft

now that we have the length, we can plug it in with one of the equations to solve for w:

w+125=345

width=$\boxed{220}$ft