Answer:
Zx = 131
Step-by-step explanation:
The angles will equal to 180, so therefore you will do 180-49=Zx
Hello,
6b) (i) As you can see, in the first year the price drops from 27,000 to 17,000. (Look at year 0-1 on the x axis). To find the percentage drop, find the difference between the two values and divide it over the initial value of 27,000.
So, the percentage drop in the first year is:
(27000-17000) / (27000) = 0.37, or a 37% drop
The answer is 37%.
(ii) For this question, we basically have the same process as the previous question except for the second year.
From year 1 to year 2, the value starts at 17,000 and ends at 15,000.
To find the percentage drop, we do:
(17000 - 15000) / (17000) = 0.118 ≈ 0.12, or a 12% drop
The answer is 12%.
6c) To find the percentage depreciation over the first 5 years, we look at the initial value (x = 0) and the value after 5 years (x = 5), and use these values in the same percentage formula we have been using.
The initial value of the car is 27,000, and after 5 years the value is 8,000.
This is a percentage drop of (27000 - 8000) / (27000) = 0.70, or a 70% drop.
The answer is 70%.
Hope this helps!
Answer:
Population of dice mice after one year = 2000
Population of wood rats after one year = 1000
The population of deer mice is growing faster than the popular of wood rats
Step-by-step explanation:
The expression for population = dN/dt = rN
Upon integration N = rN²/2
Therefore for population N = 200 and r= 0.1
N after one year = (0.1 x 200²)/ 2 = 2000
Therefore for population N = 100 and r= 0.2
N after one year = (0.2 x 100²)/ 2 = 1000
Hence the population of deer mice is growing faster than the popular of wood rats
Answer:
Step-by-step explanation:
4) ΔSTW ≅ ΔBFN . So, corresponding parts of congruent triangles are congruent.
a) BN = SW d) m∠W = m∠N
BN = 9 cm m∠W = 82°
b) TW = FN e) m∠B = m∠S
TW = 14 cm m∠B = 67°
c) BF = ST f) m∠B + m∠N + m∠F = 180°
BF = 17 cm 67 + 82 + m∠F = 180
149 + m∠F = 180
m∠F = 180 - 149
m∠F = 31°
5) ΔUVW ≅ ΔTSR
UV = TS
12x - 7 = 53
12x = 53+7
12x = 60
x = 60/12
x = 5
UW =TR
3z +14 = 50
3z = 50 - 14
3z = 36
z = 36/3
z = 12
SR =VW
5y - 33 = 57
5y = 57 + 33
5y = 90
y = 90/5
y = 18
7) ΔPHS ≅ ΔCNF
∠C = ∠P
4z - 32 = 36
4z = 36 + 32
4z = 68
z = 68/4
z = 17
∠H = ∠N
6x - 29 = 115
6x = 115 + 29
6x = 144
x = 144/6
x = 24
∠P + ∠H + ∠S = 180 {Angle sum property of triangle}
36 +115 + ∠S = 180
151 + ∠S = 180
∠S = 180 - 151
∠S = 29°
∠F = ∠S
3y - 1 = 29
3y = 29 + 1
3y = 30
y = 30/3
y = 10
8) ΔDEF ≅ ΔJKL
DE = 18 ; EF = 23
DF = 9x - 23
JL= 7x- 11
DF = JL {Corresponding parts of congruent triangles}
9x - 23 = 7x - 11
9x - 7x - 23 = -11
2x - 23 = -11
2x = -11 + 23
2x = 12
x = 12/2
x = 6
JK = DE {Corresponding parts of congruent triangles}
3y - 21 = 18
3y = 18 + 21
3y = 39
y = 39/3
y = 13
