CPhill

Note that  4  = 2^2

And 64  = 2^6

So...we have 

4^x  = 64

(2^2)^x  = 2^6

2^(2x)  = 2^6

Since the bases are the same, we can solve for the exponents

2x  =  6    divide both sides by 2

x  = 3

Note that by ASA, triangles ACN  and XCN  are congruent

This means that  AN = XN  =  5

And AN + XN  = AX   = 10

So...in triangle  AXB.....M  is the midpoint of AB  and N  is the midpoint of AX

But.....if these sides are split in a 1:1 ratio....then a segment  connecting MN  must be parallel to base  BX

Therefore....we have that

AN / MN  = AX / BX

5  / MN = 10 / 14

5 / MN  = 5 / 7   

This implies that  MN   = 7

Since QR  is parallel to ST.....then segment QR  is  parallel to  base ST

This sets up the following relationship :

PQ / QS    = PR / RT

8 / 6   = PR /RT

4 / 3  = PR /RT

So....we have 7  equal parts to PT

And  PR  is 4 of them

So   PR  =  (4/7) (10.5)   =   6

3 + 2x  < -3x - 7      add 3x to both sides, subtract 3 from  both sides

5x  < -10      divide both sides by 5

x < -2

We have the following equation

( 36B + 6B + A )  + ( 4*36 + 1* 6 + B)  + (36A  + 6  + 5)  = 216A + 36 + 5*6 + 2

Simplify

43B  + 37A  + 150 +  11  = 216A + 68

43B + 37A  + 161  = 216A + 68

43B   = 179A - 93

Note....this equation will be true when B  = 2  and A  = 1

So

l A  - B  l  =   l 1   - 2   l    =  l  -1  l   =   1

(a)  We have the points  on the graph  (0, 0 )  and ( 3,6)

An equation for this  is  C = (6/3)x  = 2x

(b) We have the points on the graph (0,0)  and (4.6)

An eqution for this is  C = (6/4)x  = (3/2)x

(c)  Snce the"strawberry" graph  is "above"  the "blueberry" graph at  every point  [except at (0,0)] ....the blueberries are cheaper at any comparable weight to the strawberries

We have  two triangles to consider

First  ...triangle  AGF

Since JE  is a segment drawn parallel to base AG.....we have the following relationship :

FE / JE   = FA / AG

1 / JE  = FA / AG

1/ JE  = 5 /AG

AG = 5  (JE )        (1)

Next...we have triangle AGD

Since  HC  is a segment drawn parallel to  base AG...we have the following relationship :

DC / HC  = DA / AG

1 / HC  =  3 / AG 

AG = 3 ( HC )       (2)

Equating (2) and (3)

5 ( JE )  = 3 ( HC )

HC / JE  = 5 / 3

Here's the one I like  :

https://matrix.reshish.com/gauss-jordanElimination.php

Find the  proportionality constant,  k

y  = k / √x

4  = k / √2

4√2  = k   = √32

So

1 =  √32 / √x

√x  = √32

x  = 32

x^3 + 5x^2 - 9x  - 45

We can factor this by grouping

x^2 (x + 5)  - 9 ( x + 5)       [ ( x + 5) is the common factor] 

(x + 5) ( x^2 - 9 )

(x + 5) ( x + 3) ( x - 3)

So....only  ( x + 3)  is a factor