Answer:
Equation 1: 14t + 61s = 150
Equation 2: 14t + 25s = 78
Use Elimination (x)
(-1) (14t + 25s) = (78) (-1)
-14t - 25s = -78
14t + 61s = 150
+ -14t - 25s = -78
--------------------------
36s = 72
s = 2
Salad costs $2
Then find t
14t + 61(2) = 150
14t + 122 = 150
14t = 28
t = 2
Sandwich costs $2
We have been given in a cohort of 35 graduating students, there are three different prizes to be awarded. We are asked that in how many different ways could the prizes be awarded, if no student can receive more than one prize.
To solve this problem we will use permutations.
![_{r}^{n}\textrm{P}={_{3}^{35}\textrm{P}}](https://tex.z-dn.net/?f=_%7Br%7D%5E%7Bn%7D%5Ctextrm%7BP%7D%3D%7B_%7B3%7D%5E%7B35%7D%5Ctextrm%7BP%7D%7D)
We know that formula for permutations is given as
![_{r}^{n}\textrm{P}=\frac{n!}{(n-r)!}](https://tex.z-dn.net/?f=_%7Br%7D%5E%7Bn%7D%5Ctextrm%7BP%7D%3D%5Cfrac%7Bn%21%7D%7B%28n-r%29%21%7D)
On substituting the given values in the formula we get,
![{_{3}^{35}\textrm{P}}=\frac{35!}{(35-3)!}=\frac{35!}{32!}](https://tex.z-dn.net/?f=%7B_%7B3%7D%5E%7B35%7D%5Ctextrm%7BP%7D%7D%3D%5Cfrac%7B35%21%7D%7B%2835-3%29%21%7D%3D%5Cfrac%7B35%21%7D%7B32%21%7D)
![=\frac{35\cdot 34\cdot 33\cdot 32!}{32!}\\ \\ =35\cdot 34\cdot 33=39270](https://tex.z-dn.net/?f=%3D%5Cfrac%7B35%5Ccdot%2034%5Ccdot%2033%5Ccdot%2032%21%7D%7B32%21%7D%5C%5C%0A%5C%5C%0A%3D35%5Ccdot%2034%5Ccdot%2033%3D39270)
Therefore, there are 39270 ways in which prizes can be awarded.
Answer:
14. x=-2.5, y = -7
15. x=28 y = -20
Step-by-step explanation:
14. Let's solve this system by elimination. Multiply the first equation by -1.
-1*(4x-y)= -1*(- 3)
-4x +y = 3
Then add this to the second equation.
-4x+y = 3
6x-y= - 8
--------------
2x = -5
Divide each side by 2
x = -2.5
We still need to find y
-4x+y =3
-4(-2.5) + y =3
10 +y =3
Subtract 10 from each side.
y = 3-10
y = -7
15.I will again use elimination to solve this system, because using substitution will give me fractions which are harder to work with. I will elimiate the y variable. Multiply the first equation by 11
11(
5x+6y)= 11*20
55x+66y = 220
Multiply the second equation by -6
-6(9x+11y)=32*(-6)
-54x-66y = -192
Add the modified equations together.
55x+66y = 220
-54x-66y = -192
---------------------------
x = 28
We still need to solve for y
5x+6y = 20
5*28 + 6y =20
140 + 6y = 20
Subtract 140 from each side
6y = -120
Divide by 6
y = -20
All circles are similar because all circle have same shape.
However Two circles are congruent , if and only if their radii are congruent.
In our question we are asked when are all circles similar given if their radii are congruent.
So we can say this statement is false because no matter the radii are congruent or not , two or more circles are always similar because of their same shape no matter what the measure of their radii is. Only if we are asked when they are congruent, we will consider the radii part.
Answer is false.
Answer:
a) 5/-6, - 5/6
b) 7/-2, -7/2
Step-by-step explanation: