Answer:
274560
Step-by-step explanation:
We can "choose" one of the 66 athletes for the gold medal. That athlete can't win any other medals, so there are 65 athletes left.
We can then "choose" one of the 65 athletes left for the silver medal. That athlete can't win any other medals, so there are 64 athletes left.
We can then "choose" one of the 64 athletes left for the bronze medal.
That leaves us with 66 possible choices * 65 possible choices * 64 possible choices=274560 possible choices
Answer:I think a is:78.3 since it would be the same and since 0 doesn't have a value.
I think b is: 139.5 because all you are doing is subtracting 287.1-147.6
Answer:
1.5225 gumbo squared
Step-by-step explanation:
First we know that 1 gumbo is equal to 8.5 lollygams.
To know the number of gumbo squared in 110 lollygam squared, we need to know the relation between 1 gumbo squared and 1 lollygam squared, and we do that making the square of our relation between gumbo and lollygams:
1 gumbo -> 8.5 lollygams
1 gumbo squared -> (8.5)^2 lollygams squared = 72.25 lollygams squared
Now, to know the number of gumbo squared in 110 lollygam squared, we just need to divide 110 by 72.25:
1 gumbo squared -> 72.25 lollygams squared
x gumbo squared -> 110 lollygams squared
x = 110/72.25 = 1.5225 gumbo squared
Answer:
vertex: (2,-18)
Step-by-step explanation:
y = ax² + bx + c
(-1,0) : a - b + c = 0 ...(1)
(5,0) : 25a +5b +c = 0 ...(2)
(0,-10): 0a + 0b + c = -10 c=-10
(1) x5: 5a - 5b + 5c = 0 ...(3)
(2)+(3): 30a + 6c = 0 30a = -6c = 60 a = 2
(1): 2 - b -10 = 0 b = -8
Equation: y = 2x² - 8x -10 = 2 (x² -4x +4) - 18 = 2(x-2)² -18
equation: y = a(x-h)²+k (h,k): vertex
vertex: (2,-18)
To find the hypotenuse of a right triangle you can use the Pythagorean theorem. (a² + b² = c²) Using the measurements provided we can solve like this.
Both a² & b² = 7²
7 x 7 = 49
49 + 49 = 98
Now we need to find the square root of 98 (c²) to get our answer.
9.9 x 9.9 = 98.01 or about 98
The measurements for the tirangle are 7 units, 7 units, and 9.9 units.
Hope this helps!
