Given: segment ac is perpendicular to segment bd prove: δacb ~ δecd reflect δecd over segment ac. this establishes that ∠acb ≅ ∠e'c'd'. then, this establishes that therefore, δacb ~ δecd by the aa similarity postulate. translate point e' to point a; ∠e'd'c' ≅ ∠abc translate point d' to point a; ∠e'd'c' ≅ ∠bac translate point d' to point b; ∠d'e'c' ≅ ∠bac translate point e' to point a; ∠d'e'c' ≅ ∠bac

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

75

step-by-step explanation:

you do x-29=46 (the number of cantaloupes sold)

in order to find x, you have to find what minus 29 equals 46, meaning you'd have to add 29 to 46.

29+46=75

if you replace x with 75, the equation looks like:

75-29=46

x=75

a precentage of 6th graders. the population is all 6th graders. let me know if i'm wrong, i hope this ! : )

step-by-step explanation:

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -