First, find the x-intercept.
The x-intercept will be at the point (x, 0) where x is any real number. If we substitute the x-coordinate and the y-coordinate for the x and y variables in the equation, we can solve for x.
5y + 3x = 15
5(0) + 3x = 15
3x + 15
x = 5
We found the x-intercept now find the y-intercept with the same process.
The y-intercept will be at the point (0, y) where y is any real number.
5y + 3x = 15
5y + 3(0) = 15
5y = 15
y = 3
So, the x-intercept is (5, 0) and the y-intercept is (0, 3)
6 x 13/3=
78/3
divide top and bottom by 3
26/1
26
Answer:
In words the answer is between t=0 and t=2.
In interval notation the answer is (0,2)
In inequality notation the answer is 0<t<2
Big note: You should make sure the function I use what you meant.
Step-by-step explanation:
I hope the function is h(t)=-16t^2+32t because that is how I'm going to interpret it.
So if we can find when the ball is on the ground or has hit the ground (this is when h=0) then we can find when it is in the air which is between those 2 numbers.
0=-16t^2+32t
0=-16t(t-2)
So at t=0 and t=2
So the ball is in the air between t=0 and t=2
Interval notation (0,2)
Inequality notation 0<t<2
The answer is c. Hope this helps
Simplify the ratio 8 to 2 to be 4 to 1
Now compare 4 to 1 and 10 to 1
4 is smaller than 10 so the ratio 8 to 2 is smaller.
