The domain is the limits of the function. Since time doesnt go negative, start with 0. At time 0, the height is
-16(0)^2 + 144 = 144ft
Then, solve for t to find the upper limit for t, which is when the height is zero (since you're dropping the object).
-16t^2 + 144 = 0
-16t^2 = -144
t^2 = 9
t = sqrt(9)
t = 3
The domain is 0 to 3 seconds.
The answer in simplest form would be 1/6
An equivalent expression would be 6(8x - 5)
The true statement about the triangle is (a) b^2 + c^2 > a^2
<h3>How to determine the true inequality?</h3>
The sides are given as:
a, b and c
The angle opposite of side length a is an acute angle
The above means that:
The side a is the longest side of the triangle.
The Pythagoras theorem states that:
a^2 = b^2 + c^2
Since the triangle is not a right triangle, and the angle opposite a is acute.
Then it means that the square of a is less than the sum of squares of other sides.
This gives
a^2 < b^2 + c^2
Rewrite as:
b^2 + c^2 > a^2
Hence, the true statement about the triangle is (a) b^2 + c^2 > a^2
Read more about triangles at:
brainly.com/question/2515964
#SPJ1
Answer:
proof below
Step-by-step explanation:
Remember that a number is even if it is expressed so n = 2k. It is odd if it is in the form 2k + 1 (k is just an integer)
Let's say we have to odd numbers, 2a + 1, and 2b + 1. We are after the sum of their squares, so we have (2a + 1)^2 + (2b + 1)^2. Now let's expand this;
(2a + 1)^2 + (2b + 1)^2 = 4a^2 + 4a + 4b + 4b^2 + 4b + 2
= 2(2a^2 + 2a + 2b^2 + 2b + 1)
Now the sum in the parenthesis, 2a^2 + 2a + 2b^2 + 2b + 1, is just another integer, which we can pose as k. Remember that 2 times any random integer, either odd or even, is always even. Therefore the sum of the squares of any two odd numbers is always even.
