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Evgen [1.6K]
2 years ago
9

The graph shows two lines, A and B. A coordinate plane is shown. Two lines are graphed. Line A has equation y equals negative 3

x plus 15. Line B has equation one half x plus 2. Based on the graph, which statement is correct about the solution to the system of equations for lines A and B? (4 points) a (4, 4) is the solution to line A but not to line B. b (4, 4) is the solution to both lines A and B. c (5, 1) is the solution to line B but not to line A. d (0, 2) is the solution to both lines A and B. DUE IN TWOOOOOOOOO MINNNNNNNNNNNNNNNNNN
Mathematics
1 answer:
AVprozaik [17]2 years ago
5 0

Answer:

Step-by-step explanation:

JUST HERE FOR THE POINTS

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5 0
2 years ago
What is the solution to the following system of equations?
ICE Princess25 [194]

Answer: B. (-6, 5)

Step-by-step explanation:

3 0
3 years ago
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What is the answer Simplify. 1/4(1-2/3 squared+1/3
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Steps: 1/4(1-(2/3^2+1/3)
3 0
3 years ago
A growth factor is a value that is greater than 1?
amm1812

Answer:

No, it is a value that is greater than 0

Step-by-step explanation:

A growth factor is indeed a value greater than 1 except it can be a decimal below 1 to. Therefore, a growth value is any POSITIVE number.

6 0
2 years ago
Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.
Elden [556K]

Answer:

The answer is "cirtical\  points \ (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})"

Step-by-step explanation:

Given:

g(x) = (x-e)^3e^{-(x-e)}

Find critical points:

g(x) = (x-e)^3e^{(e-x)}

differentiate the value with respect of x:

\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r}  \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]

critical points g'(x)=0

\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3

So,

The critical points of (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})

7 0
3 years ago
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