Answer:
(-3, 5)
Step-by-step explanation:
parallelograms and relatively even shapes, so how ever far apart the bottom two vertices are apart is how far apart you would space the top ones. simply add the x coordinates absolute values (meaning ignore negative values for now) 6+0=0. 6 Units away from 3( top vertex is (3,7)) would be -3. for the y coordinates you gotta look at y coordinates of the top vertex and bottom right vertex. they're ten units apart. go ten up from -5( because the bottom left vertex is (-6,-5) and you get 5. so overall (-3,5) would be the last vertex
Answer:
20 possible arrangements
Step-by-step explanation:
A: Alice
B: Bob
C: Cindy
X: other student
All the possible arrangements are:
When Bob is in the first place
B A C X X
B A X C X
B A X X C
B X A C X
B X A X C
B X X A C
6 possible arrangements
When Bob is in the second place (notice that neither Alice nor Cindy can be before Bob)
X B A C X
X B A X C
X B X A C
3 possible arrangements
When Bob is in the third place
X X B A C
Only 1 possible arrangement.
Total arrangements when Bob is before Cindy: 10
If Cindy is before Bob the same situations are repeated, given another 10 arrangements. So, the total line up in which Alice is somewhere between Bob and Cindy are 20.
Domain means the values of independent variable(input) which will give defined output to the function.
Given:
The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function
Solution:
To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.
![To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq 0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq 0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)](https://tex.z-dn.net/?f=%20To%20%5C%3B%20find%20%5C%3B%20domain%3A%5C%5C%5C%5Ch%28t%29%20%5Cgeq0%5C%5C%5C%5C-16t%5E2%2B96t%20%5Cgeq%20%200%5C%5CFactoring%20%5C%3B%20-16t%20%5C%3B%20in%20%5C%3B%20the%20%5C%3B%20left%20%5C%3B%20side%20%5C%3B%20of%20%5C%3B%20the%20%5C%3B%20inequality%5C%5C%5C%5C-16t%28t-6%29%20%5Cgeq%20%200%5C%5CStep%20%5C%3B%201%3A%20Find%20%5C%3B%20Boundary%20%5C%3B%20Points%20%5C%3B%20by%20%5C%3B%20setting%20%5C%3B%20up%20%5C%3B%20above%20%5C%3B%20inequality%20%5C%3B%20to%20%5C%3B%20zero.%5C%5C%5C%5Ct%28t-6%29%3D0%5C%5CUse%20%5C%3B%20zero%20%5C%3B%20factor%20%5C%3B%20property%20%5C%3B%20to%20%5C%3B%20solve%5C%5C%5C%5Ct%3D0%20%5C%3B%20%28or%29%20%5C%3B%20t%20%3D%206%5C%5C%5C%5CStep%20%5C%3B%202%3A%20%5C%3B%20List%20%5C%3B%20the%20%5C%3B%20possible%20%20%5C%3B%20solution%20%5C%3B%20interval%20%5C%3B%20using%20%5C%3B%20boundary%20%5C%3B%20points%5C%5C%28-%20%5Cinfty%2C0%5D%2C%20%5C%3B%20%5B0%2C%206%5D%2C%20%5C%26%20%5B6%2C%20%5Cinfty%29%20)
![Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq 0\\-112 \geq 0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq 0\\80 \geq 0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq 0\\-112 \geq 0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution](https://tex.z-dn.net/?f=%20Step%20%5C%3B%203%3APick%20%5C%3B%20test%20%5C%3B%20point%20%5C%3B%20from%20%5C%3B%20each%20%5C%3B%20interval%20%5C%3B%20to%20%5C%3B%20check%20%5C%3B%20whether%20%5C%5C%5C%3B%20makes%20%5C%3B%20the%20%5C%3B%20inequality%20%5C%3B%20TRUE%20%5C%3B%20or%20%5C%3B%20FALSE%5C%5C%5C%5CWhen%20%5C%3B%20t%20%3D%20-1%5C%5C-16%28-1%29%28-1-6%29%20%5Cgeq%20%200%5C%5C-112%20%5Cgeq%20%200%20%5C%3B%20FALSE%5C%5C%28-%5Cinfty%2C%200%5D%20%5C%3B%20is%20%5C%3B%20not%20%5C%3B%20solution%5C%5CAlso%20%5C%3B%20Logically%20%5C%3B%20time%20%5C%3B%20t%20%5C%3B%20cannot%20%5C%3B%20be%20%5C%3B%20negative%5C%5C%5C%5CWhen%20%5C%3B%20t%20%3D%201%5C%5C-16%281%29%281-6%29%20%5Cgeq%20%200%5C%5C80%20%5Cgeq%20%200%20%5C%3B%20TRUE%5C%5C%20%5C%3B%20%5B0%2C%206%5D%20%5C%3B%20is%20%5C%3B%20a%20%5C%3B%20solution%5C%5C%5C%5CWhen%20%5C%3B%20t%20%3D%207%5C%5C-16%287%29%287-6%29%20%5Cgeq%20%200%5C%5C-112%20%5Cgeq%20%200%20%5C%3B%20FALSE%5C%5C%20%5C%3B%20%5B6%2C%20-%5Cinfty%29%20%5C%3B%20is%20%5C%3B%20not%20%5C%3B%20solution%20)
Conclusion:
The domain of the function is the time in between 0 to 6 seconds
The height will be positive in the above interval.
3r - 8 = 7 + 8r (make rs on 1 side, numbers on other)
5r = -15(divide by 5 now)
r = -3 (<---Answer)
Hope this helped!
Answer: 30$
Step-by-step explanation:
Equation
C=5h
c=5(6)
c=30
