We know that
the equation of a vertical parabola in vertex form
y=a*(x-h)²+k
(h,k)------> (0,5)
y=a*(x-0)²+5
y=a*x²+5
substitute the point (2,9) in the equation
9=a*(2)²+5------> 9=4*a+5-------> 4*a=9-5-----> 4*a=4-----> a=1
the equation of the vertical parabola is
y=x²+5
the equation of a horizontal parabola in vertex form
x=a*(y-k)²+h
(h,k)------> (0,5)
x=a*(y-5)²+0
x=a*(y-5)²
substitute the point (2,9) in the equation
2=a*(9-5)²------> 2=16*a------> a=1/8
the equation of the horizontal parabola is
x=(1/8)*(y-5)²
the answer isthe equation of the vertical parabola is y=x²+5
the equation of the horizontal parabola is x=(1/8)*(y-5)²
see the attached figure
Answer:
h(x) = 11/(x -4)
Step-by-step explanation:
The domain of m(x) is restricted to x≠1. That means the domain of m(n(x)) must be restricted so that n(x)≠1, or x≠4.
The only offered choice with a domain restriction x≠4 is the 3rd choice:
h(x) = 11/(x -4)
Answer: 36
Step-by-step explanation: To find the least common multiple or <em>lcm</em> of 12 and 18, begin by listing the first few multiples of each number.
<em>Multiples of 12</em>
1 × 12 = 12
2 × 12 = 24
3 × 12 = 36
Notice that we skipped 0 × 12 in our list of multiples. That's because 0 × 6 is 0 and our least common multiple can not be 0.
When listing the multiples of 18, it's a good idea to keep an eye on the list of multiples for 12 so that we will notice when we find a least common multiple.
<em>Multiples of 18</em>
1 × 18 = 18 ← is not multiple of 12
2 × 18 = 36 ← is a multiple of 12
Notice that we can stop here because all other multiples that we find will be greater 36. So the least common multiple of 12 and 18 is 36.
It will take the train 14 hours to reach miles,
I got this by dividing 35 by 490.
Answer: 1.704 x 10^9
Step-by-step explanation: Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the
10
. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.