Answer:
1. A
2. B
3. D
Step-by-step explanation:
Use PEMDAS to solve all
1.
8^2 - 6*2
8*8 - 6*2
64 - 12 = 52
2.
4^2 * 3 + 4 * 2
4 * 4 * 3 + 4 * 2
16 * 3 + 8
48 + 8 = 56
3.
9^2 - 2(5+3)
9 * 9 - 2(8)
81 - 16 =65
the standard form for a horizontal ellipse is
X^2/a^2 + y^2/b^2 = 1
Substitute 54 for b and and use (8,18) as the
point to find a
x=8
y=18
8^2/a^2 + 18^2/54^2 =1
64/a^2 + 324/2916 = 1
324/2916 reduces to 1/9
64/a^2 + 1/9 = 1
64/a^2= 1-1/9
64/a^2 = 8/9
64*9/8 = a^2
576/8 = 72
A^2 = 72
A = square root(72) = 8.485
So formula
would be x^2/72 + y^2/2916 =1
Which of the following smallest values
12 three credit classes and two 2 credit classes
EXPLINATION:
12x3=36
2x2=4
And when you add you’re products together you get you get 40
Answer:
Do what the 6th grader said
Step-by-step explanation:
They smart
