B] The rule represents the reflection over y-axis is given by:
(x,y)⇒(-x,y)
c]Using this rule, the image would be located onto the pre-image when reflected again. This implies that it will be mapped onto the original point.
d] When the new pentagon is rotated 90° clockwise, the new coordinates will be as follows:
the rule for this rotation is given by:
(x,y)⇒(-y,x)
thus using the above rule new coordinates will be:
A"(-1,8), B"(-3,9), C"(-2,6), D"(-5,3) and E"(-1,4). The new figure is shown in the graph.
The Answer is -6+3=-3 because the little dot thing starts at -6 but it’s going to the right so when it goes to the right that means your going to have to add so your going to add 3 and if you add -6+3 you will get negative three ! Hope this helps !
A. Here's a fifth degree polynomial,
f(x) = x⁵ + 4x⁴ - 14x² + 9
It's in standard form, each term a constant coefficient times a whole number power of x (including the constant, which we can think of as the coefficient on x⁰=1), with the terms sorted from highest degree (highest power on x) to lowest.
B. The closure of addition wrt polynomials just means when we add two polynomials we get another polynomial.
f(x) = x⁵ + 4x⁴ - 14x² + 9
g(x) = -2x⁴ + x
f(x)+g(x) = x⁵ + 2x⁴ - 14x² + x + 9
We added two polynomials, we got another one, that's all closure is.
True because b will always be +1 than a
Answer:
1) h = -1/2t^2 +10t
2) h = -1/2(t -10)^2 +72
3) domain: [0, 20]; range: [0, 50]
Step-by-step explanation:
1.) I find it easiest to start with the vertex form when the vertex is given. The equation of the presumed parabolic path for Firework 1 is ...
h = a(t -10)^2 +50
To find the value of "a", we must use another point on the graph. (0, 0) works nicely:
0 = a(0 -10)^2 +50
-100a = 50 . . . . . . subtract 100a
a = -1/2 . . . . . . . . . divide by -100
Then the standard-form equation is ...
h = (-1/2)(t^2 -20t +100) +50
h = -1/2t^2 +10t
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2.) The path of Firework 2 is translated upward by 22 units from that of Firework 1.
h = -1/2(t -10)^2 +72
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3.) The horizontal extent of the graph for Firework 1 is ...
domain: 0 ≤ t ≤ 20
The vertical extent of the graph for Firework 1 is ...
range: 0 ≤ h ≤ 50