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Igoryamba
3 years ago
5

How can you tell the zeros of this function by looking at the graph? What are the zeros of the function?

Mathematics
2 answers:
Harrizon [31]3 years ago
7 0
Zero points are these points, where your graph cross x-axis (vertical). And zeros are x values for these points
alexgriva [62]3 years ago
3 0

Answer:C

Step-by-step explanation: Zeros occur where the graph intercepts the x-axis. On this graph the zeros are -1,1, and 3

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77julia77 [94]
the answer is x=4.
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3 years ago
Betsy was in a car accident. Her monthly insurance payment of $225 will now be increased by 25%. How much will her monthly payme
sleet_krkn [62]

Answer:

281.25 that is the answer

5 0
2 years ago
Question 19 of 25
Ronch [10]

The next step in the series will be Option B

<h3>What is a Series ?</h3>

A series is the sequence of object in pattern.

From the given figure it can be seen that the red and blue colour semicircle are shifting alternately , and the number series is increasing by 1

Therefore the next step is red on the right side , blue on the left side and no. 4 written on blue

Therefore Option B is the right answer.

To know about Series

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4 0
2 years ago
PLEASE HELP ME with this question!!
mezya [45]

Answer:

b) x=13

Step-by-step explanation:

first divide 10 by 2 to get the base of one side of the triangle

the answer will be 5

Then using the Pythagorean theorem

a2+b2=c2

12^2+5^2=169

the square root of 169 will be 13

6 0
3 years ago
Let V be the event that a computer contains a virus, and let W be the event that a computer contains a worm. Suppose P(V) = 0.17
ArbitrLikvidat [17]

Answer: 0.82

Step-by-step explanation:

The probability of the computer not containing neither a virus nor a worm is expressed as P(V^{C}∩W^{C}) , where P(V^{C}) is the probability that the event V doesn't happen and P(W^{C}) is the probability that the event W doesn't happen.

P(V^{C})= 1-P(V) = 1-0.17 = 0.83

P(W^{C})=1-P(W) = 1-0.05 = 0.95

Since V^{C} and W^{C} aren't mutually exclusive events, then:

P(V^{C}∪W^{C}) = P(V^{C}) + P(W^{C}) - P(V^{C}∩W^{C})

Isolating the probability that interests us:

P(V^{C}∩W^{C})= P(V^{C}) + P(W^{C}) - P(V^{C}∪W^{C})

Where P(V^{C}∪W^{C}) = 1 - 0.04 = 0.96

Finally:

P(V^{C}∩W^{C}) = 0.83 + 0.95 - 0.96 = 0.82

5 0
3 years ago
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