Identify the means and the extremes in the proportion.
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Identify the means and the extremes in the proportion. . Means: y & x Extremes: w & z. w = y x z. Identify the means and the extremes in the proportion. . Means: 6 + y & 12w Extremes: 9 & 5. 9 = 12w 6 + y 5. - PowerPoint PPT Presentation
Transcript of Identify the means and the extremes in the proportion.
Identify the means and the extremes in the proportion.
w = y x z
Means: y & xExtremes: w & z
Identify the means and the extremes in the proportion.
9 = 12w 6 + y 5
Means: 6 + y & 12wExtremes: 9 & 5
Identify the means and the extremes in the proportion.
q + 3 = 2 5m q
Means: 5m & 2Extremes: q + 3 & q
Identify the means and the extremes in the proportion.
m = T a h
Means: a & TExtremes: m & h
Factor completely.
x2 – 225
(x + 15)(x – 15)
Factor completely.
5x2 – 125y4
5(x2 – 25y4)5(x + 5y2)(x – 5y2)
Factor completely.
12x13 – 75x
3x(4x12 – 25)3x(2x6 + 5)(2x6 – 5)
Factor completely.
x2 + 12x + 32
(x + 4)(x + 8)
Factor completely.
x2 – 12x + 32
(x – 4)(x – 8)
Factor completely.
x2 – 4x + 32
(x + 4)(x – 8)
Factor completely.
x2 + 14x + 45
(x + 9)(x + 5)
Factor completely.
x2 + 15x + 36
(x + 12)(x + 3)
Factor completely.
8x2 + 2x – 21
8 •-21 = -168 14 & -12
8x2 + 14x – 12x – 21 2x(4x + 7) – 3(4x + 7) (4x + 7)(2x – 3)
Factor completely.
12x2 + 53x + 20
12 •20 = 240 48 & 5
12x2 + 48x + 5x + 20 12x(x + 4) + 5(x + 4) (x + 4)(12x + 5)
Factor completely.
12x2 – 16x – 11
12 •-11 = -132 -22 & 6
12x2 – 22x + 6x – 11 2x(6x – 11) + 1(6x – 11) (2x + 1)(6x – 11)
Factor completely.
32x2 – 36x + 9
32 • 9 = 288 -24 & -12
32x2 – 24x – 12x + 9 8x(4x – 3) – 3(4x – 3) (8x – 3)(4x – 3)
Factor completely.
x2 – 15x + 36
(x – 12)(x – 3)
Factor completely.
x2 + 4x – 45
(x – 9)(x + 5)
Factor completely.
2x2 – 28x + 90
2(x2 – 14x + 45)2(x – 9)(x – 5)
Factor completely.
3x2 – 27x – 108
3(x2 – 9x – 36)3(x – 12)(x + 3)
16x
12
Find the missing side of the triangle.
122 + 162 = x2
144 + 256 = x2
400 = x2
20 = x
3x
2
Find the missing side of the triangle.
22 + 32 = x2
4 + 9= x2
13 = x2
3.6 ≈ x
10x
18
Find the missing side of the triangle.
102 + 182 = x2
100 + 324= x2
424 = x2
20.6 ≈ x
1219
x
Find the missing side of the triangle.
122 + x2 = 192
144 + x2 = 361 x2 = 21714.7 ≈ x
x70
30
Find the missing side of the triangle.
302 + x2 = 702
900 + x2 = 4900 x2 = 400063.2 ≈ x
x25
7
Find the missing side of the triangle.
72 + x2 = 252
49 + x2 = 625 x2 = 57624 = x
Do the following lengths form a right triangle?
72 + 242 ? 252
49 + 576 ? 625 625 = 625 yes!
7, 24, 25
Do the following lengths form a right triangle?
52 + 122 ? 132
25 + 144 ? 169 169 = 169 yes!
5, 13, 12
Do the following lengths form a right triangle?
92 + 122 ? 152
81 + 144 ? 225 225 = 225 yes!
9, 12, 15
Do the following lengths form a right triangle?
122 + 182 ? 202
144 + 324 ? 400 468 ≠ 400 no!
12, 20, 18
Do the following lengths form a right triangle?
172 + 182 ? 192
289 + 324 ? 361 613 ≠ 361 no!
19, 17, 18
Do the following lengths form a right triangle?
22 + 72 ? 82
4 + 49 ? 64 53 ≠ 64 no!
2, 7, 8
A C
BD
∆ABC ~ ∆______ ~ ∆ _______
DBA DAC
B Y
US
∆YUB ~ ∆______ ~ ∆ _______
YBS BUS
Q R
PM
∆PQR ~ ∆______ ~ ∆ _______
PMQ QMR
H J
GI
∆GHJ ~ ∆______ ~ ∆ _______
GIH HIJ
∆ABC ~ ∆DEF. Find the value of x.
B A
C
3x + 3
4xD E
F2127
30
4x = 30 3x + 3 27
90x + 90 = 108x 90 = 18x x = 5
∆GHI ~ ∆JKL. Find the value of x.
L K
J
3x – 20
2x + 5G I
H98
10
2x + 5 = 9 3x – 20 8
16x + 40 = 27x – 180 220 = 11x x = 20
∆MNO ~ ∆PQR. Find the value of x.
N O
M
5x – 6 6x +
1
Q R
P28.546
.5
34.5
6x + 1 = 46.5 5x – 6 28.5
232.5x – 279 = 171x + 28.5 61.5x = 307.5 x = 5
