Have you calculated the temperatures?
Answer:
the object is in the air on the time interval (0.24 sec, 6.51 sec)
Step-by-step explanation:
The object is 'in the air' for all t such that h> 0. We need to find the roots of h = -16t^2 + 108t - 25 = 0. From the graph we see that both t values are positive. Once we find them, we subtract the smaller t from the larger t, which results in the length of time the object is in the air.
Use the quadratic formula to find the roots of h(t). The coefficients of t are {-16, 108, -25}, and so the discriminant b^2 - 4ac is
108² - 4(-16)(-25) = 11664 - 1600 = 10064, whose square root is 100.32.
Then the quadratic formula x = (-b ± √[b² - 4ac)/(2a) becomes
-108 ± 100.32 108 ± 100.32
t = ---------------------- = --------------------- = 3.375 ± 3.135
2(-16) 32
or t = 6.51 or t = 0.24 (both times expressed in seconds).
So, again, the object is in the air on the time interval (0.24 sec, 6.51 sec)
Step-by-step explanation:
again, 2 unknowns.
x = number of 2- points baskets
y = number of 3- points baskets
we know they hit the basket 37 times in a so far unknown mixture of 2-point and 3- point throws.
x + y = 37
which gives us e.g.
x = 37 - y
and we know that with this unknown mixture they scored 80 points.
so,
2x + 3y = 80
as every successful 2-points throw scores 2 points, and every successful 3-points throw scores 3 points.
so, again our 2 equations :
x = 37 - y
2x + 3y = 80
remember, we prepared the first equation so that it gives us already an identity expressing one variable by the other. and that we use in the second equation :
2×(37 - y) + 3y = 80
74 - 2y + 3y = 80
74 + y = 80
y = 6
and from
x = 37 - y
we get
x = 37 - 6 = 31
so, they had 31 2-points throws and 6 3-points throws.
Answer:
5(2 + y)
Step-by-step explanation:
The sum of 2 and y is 2 + y and 5 times this sum is
5(2 + y)
If the math adds up the submarine will travel .6 km in .25 hours.
