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].
Answer:
y ≥ -x +2
Step-by-step explanation:
The solid line has a slope of -1 and a y-intercept of 2, so its equation in slope-intercept form is ...
y = -x +2
The shaded area is above this line, and the line is part of the solution set, so we want an inequality that has "y" and the comparison symbol in this order: "y ≥" or "≤ y".
We already have an equation with "y" on the left, above, so we just need to introduce the comparison symbol:
y ≥ -x +2
Another way to write this is ...
x + y ≥ 2
Answer:
The length of the hall way
the weight of the wombat
Answer:
x = 145°
Step-by-step explanation:
We know that y = 98°, so we can find ∠DBC:
180°-y = ∠DBC
180°-98° = 82°
We also know that z = 63°, therefore we can apply <u>exterior angle of triangles</u> to find x.
∠DBC + z = x
82°+63° = x
x = 145°
Answer:
a,d
Step-by-step explanation:
