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Doss [256]
4 years ago
11

Compare the function ƒ(x) = –x2 + 4x – 5 and the function g(x), whose graph is shown. Which function has a greater absolute maxi

mum (vertex)? Question 16 options: A) There isn't enough information given. B) g(x) and ƒ(x) have equal absolute maximums. C) g(x) D) ƒ(x)

Mathematics
1 answer:
hram777 [196]4 years ago
8 0

Answer:

Answer C:  g(x)

Step-by-step explanation:

I used a graphing calculator to graph f(x) = -x^2 + 4x - 5, and by doing so I immedately saw that the vertex of f(x) is at (2, -1).

The absolute max of g(x) is approximately (3.25, 6.1).

The absolute max of f(x) is approximately (2, -1).

Since the y-coordinate of the absolute maximum of g(x) is greater than the y-coordinate of the absolute maximum of f(x), we conclude that Answer C is correct:  g(x) has the greater absolute maximum

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Answer:

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7 0
3 years ago
Part A: Solve –vp + 40 < 95 for v. Show your work. Part B: Solve 8w – 3r = 45 for r. Show your work.
Darya [45]

Part a

-vp + 40<95

-vp +40-40 <95-40

-vp < 55

-vp/p < 55/p

-v< 55/p

(-1)(-v) < (-1) (55/p)

v< -55/p


Part b

8w-3r=45

8w-8w-3r = 45-8w

-3r = 45-8w

-3r/-3 = 45-8w/-3


8 0
3 years ago
The rule r y-axis R 0,90^ (x,y) is applied to triangle ABC . Which triangle shows the final image?
NemiM [27]

Answer:

1

Step-by-step explanation:

Given  triangle ABC with vertices at points A(2,-4), B(4,-4) and C(4,-2).

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(x,y)\rightarrow (-y,x),

then

A(2,-4)\rightarrow A'(4,2)\\ \\B(4,-4)\rightarrow B'(4,4)\\ \\C(4,-2)\rightarrow C'(2,4)

2. Reflect triangle A'B'C' across the y-axis. This reflection has the rule

(x,y)\rightarrow (-x,y),

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A'(4,2)\rightarrow A''(-4,2)\\ \\B'(4,4)\rightarrow B''(-4,4)\\ \\C'(2,4)\rightarrow C''(-2,4)

These are vertices of triangle 1

5 0
3 years ago
Read 2 more answers
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Answer:

Incorrect

Step-by-step explanation:

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3 years ago
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Answer:

1, 2, 4, 5, 10, 20

Step-by-step explanation:

they all can go into 20

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