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3 years ago

Given the following linear function sketch the graph of the function and find the domain and range.

Mathematics

1 answer:

kobusy [5.1K]3 years ago

8 0

Answer:

see the procedure

Step-by-step explanation:

we have

$f(x)=-3x+7$

This is the equation of a line in slope intercept form

where

$m=-3$ ----> the slope

$b=7$ ----> the y-coordinate of y-intercept

To graph the line find out the intercepts

The y-intercept is the point (0,7) ----> given value

The x-intercept is the point (2.33,0) ---> Remember that the x-intercept is the value of x when the value of the function is equal to zero

with the points (0,7) and (2.33,0) sketch the graph

using a graphing tool

Plot the points and join to graph the line

see the attached figure

The domain is all real numbers

The range is all real numbers

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Help (ಥ‿ಥ)

mr Goodwill [35]

Answer: B

Step-by-step explanation:

6 0

2 years ago

What are the solutions to the equation<br><br>x^2+14x=-130<br><br>please do a step by step

Serjik [45]

Answer:

The answer to your question is: x1 = - 7 + 9i x2 = -7 - 9i

Step-by-step explanation:

x² + 14x = - 130

x² + 14x + (7)² = -130 + (7)²

(x + 7)² = -81

x + 7 = ± √-81

i² = -1, then x + 7 = ±√ 81i²

x + 7 = ± 9i

x1 = -7 + 9i x2 = -7 - 9i

5 0

3 years ago

A student comes to lecture at a time that is uniformly distributed between 5:09 and 5:14. Independently of the student, the prof

scZoUnD [109]

Answer:

1/2

Step-by-step explanation:

The lecture has already begun when the student arrives means one of these scenarios happen:

1) the class started at 5:10 and the student arrives at 5:11 or 5:12 or 5:13 or 5:14

2) the class started at 5:11 and the student arrives at 5:12 or 5:13 or 5:14

3) the class started at 5:12 and the student arrives at 5:13 or 5:14

Given student time of arrival is uniformly distributed, then the probability he/she arrives at 5:09 or 5:10 or 5:11 or 5:12 or 5:13 or 5:14 is 1/6.

So, the probability that the student arrives between 5:11 and 5:14 is 1/6 + 1/6 + 1/6 + 1/6 = 2/3.

The probability that the student arrives between 5:12 and 5:14 is 1/6 + 1/6 + 1/6 = 1/2.

The probability that the student arrives at 5:13 or 5:14 is 1/6 + 1/6 = 1/3.

Given class starting time is uniformly distributed, then the probability it starts at 5:10 or 5:11 or 5:12 is 1/3.

Given the two events are independent, the probability of the first scenario is: (1/3)*(2/3) = 2/9

For the second scenario: (1/3)*(1/2) = 1/6

For the third scenario: (1/3)*(1/3) = 1/9

Because all of these scenarios are mutually exclusive the total probability of one of them happen is: 2/9 + 1/6 + 1/9 = 1/2

3 0

3 years ago

At boarding school, students spend 1/4 of their time eating and sleeping, 3/5 of their time in class, and 1'12 studying. The res

steposvetlana [31]

Answer:

1/15 of their time is free time.

Step-by-step explanation:

To find the answer to this 2-step problem, we have to add all the fractions up, then find how much of that time is not part of the fraction.

So to add all the fractions, you have the equation:

1/4 + 3/5 + 1/12

The first thing we have to do is to find a common denominator.

4, 5, and 12 are each able to multiply into 60.

So we can make the equation now with a common denominator.

15/60+36/60+5/60=56/60.

We now can simplfy this.

56/60=14/15

Now that we know how long it is to do all the other activities, it tells us that the rest of the time is free time. To find the rest of the time, we have to do:

1-14/15=1/15

So 1/15 of their time is free time.

8 0

3 years ago

Four runners start at the same point; Lin, Elena, Diego, Andre. For each runner write a multiplication equation that describes t

Helen [10]

Answers:

Lin's finish point is at 205 meters

Elena's velocity is 8.92 meters per second

Diego's finish point is at -259.2 meters or 259 meters behind the start

Andre's velocity is -8.14 meters per second

I hope this helps <3

6 0

2 years ago