CPhill

Similar figures are similar in all respects.

The ratio of the lateral sides and perimeters = the ratio of square root of their areas

Then, the ratio of the sides and the perimeters is 6:5

A^2= 1/2(s-a)(s-b)(s-c)

OK  

multiply both sides by 2

2A^2 = (s-a)(s-b)(s-c)

divide both sides by (s-a)(s-b)

[2A^2]/[(s-a)(s-b)]  = s-c

Add c to both sides and subtract the expression on the left side from both sides

c = s - [2A^2]/[(s-a)(s-b)]

We could get a common denominator here, but the final result would be a little messy...this is good enough !!!

Infinite solutions (example)

3x = 3x

Note that any "x" value will make both sides equal

No solutions (example)

3x + 6 = 3x + 7

Note....adding 6 to something can never be equal to adding 7 to the same thing !!!

One solution (example)

3x = 6

Only one value of x "works"  (i.e., 2  !!!)

Hope that helps !!

(x^4-13x^2+36) /(x-2)

Let's factor the trinomial in the numerator

(x^2 - 9) (x^2 - 4) =

(x - 3)(x +3)(x- 2)(x +4)

So we have

[(x - 3)(x +3)(x- 2)(x +2)] / (x - 2)  =

[(x - 3) (x + 3) (x + 2)]

And no "long division" was necessary......

Here's a picture of the cross-section of the pyramid.....

AB = face height = 8     BC = 1/2 of the base length = 2

So the height =

√[(face height)² - (1/2 base length)²] =

√[ 8² - 2²]  = √(60) = about 7.746 ft.

No prob, radio.....I gave you points for a "good effort"......it's always easy to make mistakes on here !!!!

I think we have.......

a/5 + 15 = -(3 + 1/2)      change -(3 + 1/2)  to -3.5 

a/5 + 15 = -3.5               subtract 15 from both sides

a/5 = -15 -3.5                 simplify

a/5 = -18.5                     multiply both sides by 5

a =   (-18.5) *(5)      =    -92.5

150/40 = 15/4 = 3.75

40/16 =  5/2 = 2.5

The first thing is faster

Good job, geno......!!!!!

Another method is to just raise each side to the 30th power

This gives

10³  ??? 2¹⁰

1000 ???  (2⁵)²

1000 ???? (32)²

1000 < 1024

x = Sin C