short leg: x
long leg: 5x + 19
hypotenuse: 5x + 20
(short leg)² + (long leg)² = hypotenuse
(x)² + (5x + 19)² = (5x + 20)²
x² + 25x² + 190x + 361 = 25x² + 200x + 400
x² - 10x - 39 = 0 <em>subtracted 25x², 200x, and 400 from both sides</em>
(x - 13)(x + 3) = 0
x - 13 = 0 , x + 3 = 0
x = 13 x = -3 <em>length cannot be negative so disregard -3</em>
long leg: 5x + 19 = 5(13) + 19 = 65 + 19 = 84
hypotenuse: 5x + 20 = 5(13) + 20 = 65 + 20 = 85
Answer: short leg = 13, long leg = 84, hypotenuse = 85
Answer:
Remember, and the range of g must be in the domain of f.
a)
The domain of f(g(x)) and g(f(x)) is the set of reals.
b)
The domain of f(g(x)) is the set of nonnegative reals and the domain of g(f(x)) is the set of number such that
c)
The domain of f(g(x)) is the set of reals except the 1 and the domain of g(f(x)) is the set of reals except the 1 and -1
d)
The domain of f(g(x)) is the set of reals except 2, and the domain of g(f(x)) is the set of reals except -1.
e)
The domain of f(g(x)) is the set of nonnegative reals except -3. The domain of g(f(x)) is the set of nonnegative reals except -2.
Answer:
The probability that a randomly selected call time will be less than 30 seconds is 0.7443.
Step-by-step explanation:
We are given that the caller times at a customer service center has an exponential distribution with an average of 22 seconds.
Let X = caller times at a customer service center
The probability distribution (pdf) of the exponential distribution is given by;
Here, = exponential parameter
Now, the mean of the exponential distribution is given by;
Mean =
So, ⇒
SO, X ~ Exp()
To find the given probability we will use cumulative distribution function (cdf) of the exponential distribution, i.e;
; x > 0
Now, the probability that a randomly selected call time will be less than 30 seconds is given by = P(X < 30 seconds)
P(X < 30) =
= 1 - 0.2557
= 0.7443
Answer:
15
Step-by-step explanation:
the pattern is adding by three so 12+3=15
Let K and T be the speeds of Kirk and Thor respectively.
Note the standard definition that d=vt, distance equals velocity times time.
t=d/v
For Kirk, t=525/(T+5)
For Thor, t=490/T
Since we are told that they have traveled the same amount of time we can say t=t which leaves us:
525/(T+5)=490/T multiply both sides by T
525T/(T+5)=490 multiply both sides by T+5
525T=490T+2450 subtract 490T from both sides
35T=2450 divide both sides by 35
T=70
So Thor's average speed is 70kph, since K=T+5
K=75kph
So Kirk's average speed is 75kph