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Pavlova-9 [17]
3 years ago
14

Plz help, The graph of y = f(x) is shown below. What are all of the real solutions of f(x) = 0?

Mathematics
1 answer:
8_murik_8 [283]3 years ago
8 0

Answer:

The real solutions of f(x)=0 are: 0, 2, 5, 6

Step-by-step explanation:

We are given:

The graph of y = f(x)

We need to find all of the real solutions of f(x) = 0?

By looking at the graph we need to find the values of y when x =0

Looking at the graph, when x=0 we get

0, 2,5 and 6

So, the real solutions of f(x)=0 are: 0, 2, 5, 6

I am attaching the figure, that determines the answers.

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HURRY 20 POINTS!!!!! Given the geometric sequence an = 6(5)n − 1, which of the following values for n lies in the appropriate do
Lapatulllka [165]
<h2>Answer:</h2>

n = 1

<h2>Explanation:</h2>

Here we have the following Geometric Sequence:

a_{n} = 6(5)^{n-1}

Where:

a=6 \ \text{First term} \\ \\ r=5 \ \text{Common ratio} \\ \\ n: \ \text{Number of terms}

The first term occurs when n=1. Next the second term occurs when n=2, the third term when n=3 and so on. Therefore, the appropriate value of n that lies on the domain for n is n=1

7 0
3 years ago
Solve the following inequality: 3-5x-x^2&gt;=0
Dennis_Churaev [7]

Answer:

-5/2+-1/2√37≤x≤-5/2+1/2√37

Step-by-step explanation:

Step 1: Find the critical points

-x^2-5x+3=0

For this equation: a=-1, b=-5, c=3

−1x^2+−5x+3=0

x=−b±√b2−4ac/2a

x=−(−5)±√(−5)2−4(−1)(3)/2(-1)

x=5±√37 /−2

x=-5/2+1/2√37

Step 2: Check intervals in between critical points

x≤-5/2+1/2 √37 (Doesn't work in original inequality)

-5/2+-1/2√37≤x≤-5/2+1/2√37 (Works in original inequality)

x≥-5/2+1/2 √37 (Doesn't work in original inequality)

5 0
3 years ago
Help asap plzzz I NEED HELP !!!!!!!!
Vedmedyk [2.9K]

Answer:

1520.5 in^2

Step-by-step explanation:

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2 years ago
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Ganezh [65]

Answer:

3

Step-by-step explanation:

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What’s 5(x+12) equal
juin [17]
Multiply using the distributive property so it would be 5x+60
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