<span>For given hyperbola:
center: (0,0)
a=7 (distance from center to vertices)
a^2=49
c=9 (distance from center to vertices)
c^2=81
c^2=a^2+b^2
b^2=c^2-a^2=81-49=32
Equation of given hyperbola:
..
2: vertices (0,+/-3) foci (0,+/-6)
hyperbola has a vertical transverse axis
Its standard form of equation: , (h,k)=(x,y) coordinates of center
For given hyperbola:
center: (0,0)
a=3 (distance from center to vertices)
a^2=9
c=6 (distance from center to vertices)
c^2=36 a^2+b^2
b^2=c^2-a^2=36-9=25
Equation of given hyperbola:
</span>
The answer is $8 per hour. To get this you add the 2 1/2 and 4 1/4 and that equals 6 3/4 or 6.75. You then divide 54 by 6.75 and you get 8.
Answer:
The answer to your question is y - 0 = 40(x - 15)
Step-by-step explanation:
y = 40x + 600
Convert to y - y₁ = m(x - x₁)
Process
1.- Factor 40 in the right side of the equation
y = 40(x - 15)
2.- Give the form of the point slope form
y - 0 = 40(x - 15)
where
y₁ = 0
m = 40
x₁ = 15
Answer:
1,459
You have to divide
7,295÷5 and u get your answer
The standard form of the quadratic function f(x) = -3x^2 + 6x - 2 is f(x) = -3x^2 + 6x - 2
<h3>How to represent the
quadratic function in standard form?</h3>
The quadratic function is given as
f(x) = -3x^2 + 6x - 2
The standard form of a quadratic function is represented as:
f(x) = ax^2 + bx + c
When both equations are compared, we can see that the function f(x) = -3x^2 + 6x - 2 is already in standard form
Where
a = -3
b = 6
c = -2
Hence, the standard form of the quadratic function f(x) = -3x^2 + 6x - 2 is f(x) = -3x^2 + 6x - 2
Read more about quadratic function at
brainly.com/question/25841119
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