Answer:
Step-by-step explanation:
1a+1b(a+b-c)+1b+1c(b=c-a)+ 1a+1c(c+a-b)
1a+1b-c+1b +1c-a+1a+1c-b
1a+2b-c+a+1c-b
2a+1b+c this is the answer I think lol
Answer:
it would be $2
Step-by-step explanation:
Y = (x - 1)² + 2
y = (x² - x - x + 1) + 2
y = (x² - 2x + 1) + 2
y = x² - 2x + (1 + 2)
y = x² - 2x + 3
x² - 2x + 3 = 0
x = <u>-(-2) +/- √((-2)² 4(1)(3))</u>
2(1)
x = <u>2 +/- √(4 + 12)</u>
2
x = <u>2 +/- √(16)
</u> 2<u>
</u>x = <u>2 +/- 4
</u> 2<u>
</u>x = 1 <u>+</u> 2
x = 1 + 2 x = 1 - 2
x = 3 x = -1
y = x² - 2x + 3
y = (3)² - 2(3) + 3
y = 9 - 6 + 3
y = 3 + 3
y = 6
(x, y) = (3, 6)
or
y = x² - 2x + 3
y = (-2)² - 2(-1) + 3
y = 4 + 2 + 3
y = 6 + 3
y = 9
(x, y) = (-1, 9)
The standard equation of an ellipse is:
(x-h)²/a² + (y-k)²/b² = 1
Since the center at the origin, then h = k = 0
The height of this ellipse = 8 units and the width = 4 units, that means the major axis is b (=8) and the minor axis is a (=4)
Then the equation becomes:
x²/16 + y²/64 = 1
Answer:
3^2
Step-by-step explanation:
The applicable rules of exponents are ...
(a^b)(a^c) = a^(b+c)
(a^b)/(a^c) = a^(b-c)
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