The true statement about the function f(x) = -x² - 4x + 5 is that:
- The range of the function is all real numbers less than or equal to 9.
<h3 /><h3>What is the domain and range for the function of y = f(x)?</h3>
The domain of a function is the set of given values of input for which the function is valid and true.
The range is the dependent variable of a given set of values for which the function is defined.
- The domain of the function: f(x) = -x² - 4x + 5 are all real number from -∞ to +∞
For a parabola ax² + bx + c with the vertex 
- If a < 0, then the range is f(x) ≤
- If a > 0, then the range f(x) ≥
The vertex for an up-down facing parabola for a function y = ax² + bx + c is:
Thus,
- vertex = (-2, 9)
Range: f(x) ≤ 9
Therefore, we can conclude that the range of the function is all real numbers less than or equal to 9.
Learn more about the domain and range of a function here:
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Answer:
see attached for a graph
x = 0.5 when y = 1
Step-by-step explanation:
The graph has a y-intercept of -1, which is halfway between 0 and -2. It goes up 4 units (2 vertical grid spaces) for 1 unit to the right (1 horizontal grid space). It seems to cross y = 1 at about x = 1/2.
x = 1/2 when y = 1.
Answer:
The initial speed of the car was 80 ft/s.
Step-by-step explanation:
The deceleration is the rate at which the car speed decreases. In this case the speed of the car goes all the way down to 0 ft/s and in order to do that it travelled 50 ft. So we will call the initial speed at which the car started to brake "v_0" and use Torricelli's equation to find it. The equation is given by:
v^2 = (v_0)^2 + 2*a*S
Where v is the final speed, v_0 is the initial speed, a is the rate of acceleration and S is the space travelled. Using the values that the problem gave to us we have:
0^2 = (v_0)^2 - 2*64*50
0 = (v_0)^2 - 6400
(v_0)^2 = 6400
v_0 = sqrt(6400) = 80 ft/s
Notice that in this case "a" was negative, since the car was decelerating instead of accelerating.
The initial speed of the car was 80 ft/s.
Answers:
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Explanation:
Recall that tangent is the ratio of opposite over adjacent
tan(angle) = opposite/adjacent
So for reference angle G, we say,
tan(G) = JH/GJ = 2/1 = 2
We'll treat tan(H) in a similar fashion, but the opposite and adjacent sides swap roles. That means we'll apply the reciprocal to the result above to get 1/2 for tan(H)
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So we have this interesting property where
tan(G)*tan(H) = 2*(1/2) = 1
In general,
tan(A)*tan(B) = 1 if and only if A+B = 90 degrees
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Side note: The side sqrt(5) isn't used at all.
