Answer:
180=2x+ 24( angles opposite to equal sides)
156/2=x
x=78
Step-by-step explanation:
Answer:
(5, - 4 )
Step-by-step explanation:
Given the 2 equations
2x + 3y = - 2 → (1)
3x - y = 19 → (2)
Multiplying (2) by 3 and adding to (1) will eliminate the y- term
9x - 3y = 57 → (3)
Add (1) and (3) term by term to eliminate y
(2x + 9x) + (3y - 3y) = (- 2 + 57), that is
11x = 55 ( divide both sides by 5 )
x = 5
Substitute x = 5 into either of the 2 equations and solve for y
Substituting into (1)
2(5) + 3y = - 2
10 + 3y = - 2 ( subtract 10 from both sides )
3y = - 12 ( divide both sides by 3 )
y = - 4
Solution is (5, - 4 )
Answer:
a) 30 kangaroos in 2030
b) decreasing 8% per year
c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo
Step-by-step explanation:
We assume your equation is supposed to be ...
P(t) = 76(0.92^t)
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a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30
In the year 2030, the population of kangaroos in the province is modeled to be 30.
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b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.
The population is decreasing by 8% each year.
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c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.
P(100) = 75(0.92^100) = 76(0.0002392)
P(100) ≈ 0.0182, about 1/55th of a kangaroo
Answer:
Option D is right,.
Step-by-step explanation:
-a+4b+2c=-8 ... i
3a+b-4c=9 ... ii
b=-1 ... iii
Substitute the value of b, in i and ii
-a-4+2c =-8 or -a+2c = -4
and 3a-1-4c = 9 or 3a -4c =8
Now we have two equations in two variables
-a+2c =-4 and 3a-4c =8
a = 2c+4: substitute this in the other equatin.
3(2c+4)-4c =8 Or 2c +12 =8
2c =-4 or c = -2
Substitute in -a+2c =12
-a-2 = -4
a = -2+4 =2
a=2, b=-1 and c =-1 is the solution.