Answer:
70, I think
Step-by-step explanation:
f(10)= -10^2- 3(10)
f(10)= 100-30
f(10) = 70
Answer: 48 inches tall
Step-by-step explanation: (I just learned this) First, you square 80 and 64. 80 will be 6400 and 64 will be 4096. Then you use the Pythagorean Theorem, which is a² + b² = c². A. and B. are the legs in a triangle and C. is the hypotenuse. If you have A. and C. but not B, you square A. and C. and subtract C. from A. Then, after you get the number, you square that number, which will be√(2304), which makes 48.
The major axis of the eclipse is 12 units long
The given parameters can be represented as:
See attachment for illustration
To solve this question, we make use of the following theorem
The distance between a point and the foci sums up to the major axis
This translates to:
it takes 5 seconds to reach a height of 40 ft. and It takes 2.5 seconds to reach maximum height.
Let h(t) represent the height of the ball at time t.
Given that the height is given by:
h(t) = -16t² + 80t + 40
1) For the ball to reach a height 0f 40 ft, h(t) = 40, hence:
40 = -16t² + 80t + 40
16t² - 80t = 0
t(t - 5) = 0
t = 0 or t = 5
Hence it takes 5 seconds to reach a height of 40 ft.
2) The maximum height is at h'(t) = 0,
h'(t) = -32t + 80
-32t + 80 = 0
32t = 80
t = 2.5
It takes 2.5 seconds to reach maximum height.
Find out more at: brainly.com/question/11535666
A could be 2 while B could be 3, so -2a+3b turns into -4+9, which equals 5.
From what I know you can't really solve a a single equation with two-variables so it's just a matter of trial and error.
Just try plugging in a small number like 2 for a just to try it and you get 8b^2=72.
Divide everything by 8 to isolate b and you get that b^2=9.
Square root everything and you'll find that b=3. This is just one possible combination, I'm sure there are many more but this is obviously the one that was intended to be found.
Now that we know that a=2 and b=3 just plug them into the equation.
-2(2)+3(3)=?
-4+9=?
5
Sorry about having to use this ^ symbol, the equation maker is not working.
