Answerhold on I got you
Step-by-step explanation:
Answer: first answer choice
Step-by-step explanation:
So, the subtraction looks like this:
223
-119
_______
The reason why combining place values here is necessary is that 3 is less than 9 (so I can't just substract in the one's place without going to negative numbers)
so we combine the ones and tens places and substract the whole: 23-19 is four!
223
-119
_______
04
and then we continue
223
-119
_______
204
and that's the result!
Answer:
The lengths of Gajge’s runs have greater variability because there is a greater difference between his longest and shortest runs is the answer.
Step-by-step explanation:
given that Ty and Gajge are football players.
Carries is 15 for both and average is the same 4 for both.
But on scrutiny we find that maximum and minimum and 6 and 2 for Ty.
Hence range for Ty = 6-2 =4 (2 runs on eithre side of mean)
But for Gajge, highest is 19 and lowest is 2.
i.e. range = 19-2 =17 very much higher than that of Ty
The lengths of Gajge’s runs have greater variability because there is a greater difference between his longest and shortest runs.
Note that √(4 - t²) is defined only as long as 4 - t² ≥ 0, or -2 ≤ t ≤ 2. Then the real integral exists only if -2 ≤ x ≤ 2. (Otherwise we deal with complex numbers.)
If x = 2, then the integral corresponds to the area of a quarter-circle with radius 2. This means that the integral has a maximum value of 1/4 • π • 2² = π.
On the opposite end, if x = -2, then the integral has the same value, but the integral from 0 to -2 is equal to the negative integral from -2 to 0. So the minimum value is -π.
For all x in between, we observe that the integrand is continuous over the rest of its domain, so F(x) is continuous.
Then the range of F(x) is the interval [-π, π].
