Answer:
y=14
Step-by-step explanation:
Since this is an isosceles triangle due to opposite sides being congruent, the base angles would also be congruent.
plug in 25 in 2x and you would get 50°
Since the sum of a triangle is 180° you could set up an equation
Then, add 50+50=100+10=110
then you would then have
subtract 110 from 180, leaving you with 70
then divide and you would get 14
Check:
5(14)+10+50+50= 180
The first step is to determine the distance between the points, (1,1) and (7,9)
We would find this distance by applying the formula shown below
Distance = 10 units
If one unit is 70 meters, then the distance between both entrances is
70 * 10 = 700 meters
Answer:
Collin: about $401 thousand
Cameron: about $689 thousand
Step-by-step explanation:
A situation in which doubling time is constant is a situation that can be modeled by an exponential function. Here, you're given an exponential function, though you're not told what the variables mean. That function is ...
In this context, P0 is the initial salary, t is years, and d is the doubling time in years. The function gives P(t), the salary after t years. In this problem, the value of t we're concerned with is the difference between age 22 and age 65, that is, 43 years.
In Collin's case, we have ...
P0 = 55,000, t = 43, d = 15
so his salary at retirement is ...
P(43) = $55,000(2^(43/15)) ≈ $401,157.89
In Cameron's case, we have ...
P0 = 35,000, t = 43, d = 10
so his salary at retirement is ...
P(43) = $35,000(2^(43/10)) ≈ $689,440.87
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Sometimes we like to see these equations in a form with "e" as the base of the exponential. That form is ...
If we compare this equation to the one above, we find the growth factors to be ...
2^(t/d) = e^(kt)
Factoring out the exponent of t, we find ...
(2^(1/d))^t = (e^k)^t
That is, ...
2^(1/d) = e^k . . . . . match the bases of the exponential terms
(1/d)ln(2) = k . . . . . take the natural log of both sides
So, in Collin's case, the equation for his salary growth is
k = ln(2)/15 ≈ 0.046210
P(t) = 55,000e^(0.046210t)
and in Cameron's case, ...
k = ln(2)/10 ≈ 0.069315
P(t) = 35,000e^(0.069315t)