1.If RS=23-2x,ST=9x-5,and RT=39 find RS.
r t = r s + s t rt = rs + st r t = rs + s t
rt = 39
rs = 23 – 2x
st = 9x-5
39 = ( 23 – 2 x ) + ( 9 x − 5 ) 39 = (23 – 2x) + (9x - 5) 39 = ( 23–2 x ) + ( 9 x − 5 )
x = 3 x = 3 x = 3
r s = 23 − 2 ∗ 3 rs = 23 - 2 * 3 rs = 23 − 2 ∗ 3
RS = 17
2.If LN=6x-5,LM=x+7, and MN=3x+20 find MN
6 ∗ x − 5 = x + 7 + 3 ∗ x + 20 6*x - 5 = x + 7 + 3*x + 20 6 ∗ x − 5 = x + 7 + 3 ∗ x + 20
6 x − 4 x = 32 6x - 4x = 32 6 x − 4 x = 32
x = 16 x = 16 x = 16
M N = 3 ∗ 16 + 20 MN = 3*16+20 MN = 3 ∗ 16 + 20
MN = 68
3.Find X Round tithe nearest hundredth.
X(-9,2) and Y(5,-4)
d = ( 5 − ( − 9 ) ) 2 + ( − 4 − 2 ) 2 d = \sqrt{ (5-(-9))^2 + (-4-2)^2} d = ( 5 − ( − 9 ) ) 2 + ( − 4 − 2 ) 2
answer: 15.23
4.Find the midpoint of AB if A(-3,8) and B(-7,-6)
m i d p o i n t = ( ( − 3 − 7 ) / 2 ; ( 8 − 6 ) / 2 ) midpoint = ((-3-7)/2; (8-6)/2) mi d p o in t = (( − 3 − 7 ) /2 ; ( 8 − 6 ) /2 )
answer: (-5 , 1)
5.If CE=10x+18, DE=7x-1 and BC=9x-2, find AB
C E = 10 x + 18 CE=10x+18 CE = 10 x + 18
D E = 7 x − 1 DE=7x-1 D E = 7 x − 1
D is midpoint
D E = 1 / 2 C E DE = 1/2CE D E = 1/2 CE
7 ∗ x − 1 = 1 / 2 ∗ ( 10 ∗ x + 18 ) 7*x - 1 = 1/2*(10*x+18) 7 ∗ x − 1 = 1/2 ∗ ( 10 ∗ x + 18 )
X=5
C E = 10 ∗ 5 + 18 = 68 = A C CE = 10*5+18 = 68 = AC CE = 10 ∗ 5 + 18 = 68 = A C
B C = 9 ∗ x − 2 = 9 ∗ 5 − 2 = 43 BC = 9*x-2 = 9*5-2 = 43 BC = 9 ∗ x − 2 = 9 ∗ 5 − 2 = 43
AB = AC - BC = 68-43 = 25
AB=25
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