The equation of the line perpendicular to x = -4 that passes through the point (-2,5) is y = 5 .
In the question ,
it is given that
the required line is perpendicular to x = -4
the slope of x = -4
x + 4 = 0
0.y = x + 4
slope = 1/0
so the slope of the perpendicular line 0 .
the equation of the perpendicular line passing through (-2,5) and slope as 0 is
(y - 5) 0*(x + 2)
y -5 = 0
y = 5
Therefore , The equation of the line perpendicular to x = -4 that passes through the point (-2,5) is y =5 .
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Answer:
a.
b.
c.
Step-by-step explanation:
In a, the -1 means inverse of f(x). To find the inverse, you rewrite the equation in terms of x and y. You replace the x with y and y with x. Then you solve for y.
In b, all you have to do is plug in -9 into f(x).
In c, you plug in -9 into inverse function f.
i don't know specifically what you are looking for but:
Direction: Opens Up
Vertex: (1, -1)
Axis of Symmetry: x = 1
and i attached the graph
Answer: x = √(((-F - 4)^2)/16 - 1/2) + (F + 4)/4 or x = (F + 4)/4 - √(((-F - 4)^2)/16 - 1/2)
Step-by-step explanation:
Solve for x:
F x = 2 x^2 - 4 x + 1
Subtract 2 x^2 - 4 x + 1 from both sides:
-2 x^2 + F x + 4 x - 1 = 0
Collect in terms of x:
-1 + x (F + 4) - 2 x^2 = 0
Divide both sides by -2:
1/2 + 1/2 x (-F - 4) + x^2 = 0
Subtract 1/2 from both sides:
1/2 x (-F - 4) + x^2 = -1/2
Add 1/16 (-F - 4)^2 to both sides:
1/16 (-F - 4)^2 + 1/2 x (-F - 4) + x^2 = 1/16 (-F - 4)^2 - 1/2
Write the left hand side as a square:
(1/4 (-F - 4) + x)^2 = 1/16 (-F - 4)^2 - 1/2
Take the square root of both sides:
1/4 (-F - 4) + x = √(1/16 (-F - 4)^2 - 1/2) or 1/4 (-F - 4) + x = -√(1/16 (-F - 4)^2 - 1/2)
Subtract 1/4 (-F - 4) from both sides:
x = √(((-F - 4)^2)/16 - 1/2) + (F + 4)/4 or 1/4 (-F - 4) + x = -√(1/16 (-F - 4)^2 - 1/2)
Subtract 1/4 (-F - 4) from both sides:
Answer: x = √(((-F - 4)^2)/16 - 1/2) + (F + 4)/4 or x = (F + 4)/4 - √(((-F - 4)^2)/16 - 1/2)
Answer:
208.02 m
Explanation:
When the model rocket lands, y = 0; therefore, the equation becomes
Using the quadratic formula we solve for x.
Which simplifies to give
Hence, we have the solutions
since lengths cannot be negative, we choose x = 208.02 as our answer.
Hence, the model rocket lands 208.02 m from the starting point.
140 - 143 1
144 - 147 16
148 - 151 71
152 - 155 108
156 - 159 83
160 - 163 18
164 - 167 3
What is the class boundary between the sixth and seventh classes?
Answer:
Class Boundary = 1 between the sixth and seventh classes.
Step-by-step explanation:
Lengths (mm) Frequency
1. 140 - 143 1
2. 144 - 147 16
3. 148 - 151 71
4. 152 - 155 108
5. 156 - 159 83
6. 160 - 163 18
7. 164 - 167 3
The class boundary between two classes is the numerical value between the starting value of the higher class, which is 164 for the 7th class in this case, and the ending value of the class of the lower class, which is 163 for the 6th class in this case.
Therefore the class boundary between the sixth and seventh classes
= 164 - 163 = 1
Therefore Class Boundary = 1.
It can be seen that class boundary for the frequency distribution is 1.
If we take the difference between the lower limits of one class and the lower limit of the next class then we will get the class width value.
Therefore, Class width,
w = lower limit of second class - lower limit of first class
= 144 - 140
= 4
Answer:
Yes
Step-by-step explanation:
'
Yes, this F(x) is a polynomial function involving the first and null integer powers of x: x^0 and x^1.
Yes, f(x) = 7 - 2x is a polynomial function because it satisfies the general form of a polynomial, which requires that every term is in the form ax^n for nonnegative integer n.
Yes, the function f(x) = 7-2x is indeed a polynomial function. A polynomial function is a function such as a quadratic, a cubic, a quartic, and so on, for which the function is of the form f(x) = ax^n, where n is a nonnegative integer. In this case, f(x) = 7-2x can be more comprehensively written as f(x) = -2x^1 + 7x^0, which reflects the general form of a polynomial function.
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