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levacccp [35]
3 years ago
9

How do you know when a line on a graph represents a relation but not a function

Mathematics
1 answer:
Brut [27]3 years ago
6 0

Answer:

#see solution for details.

Step-by-step explanation:

-Draw a horizontal line across the subject line.

-If the drawn horizontal line intersects our subject line more than once, then the line is not function but a relation.

-A relation has more than one input

- A line is said to be a relation if it has more than one input values, and a function if it only has one input value.

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I need help ASAP please (rational exponents)
denis-greek [22]
The answer would be x^12yz^4/64
5 0
4 years ago
Secants P N and L N intersect at point N outside of the circle. Secant P N intersects the circle at point Q and secant L N inter
JulijaS [17]

Answer:

A circle is shown. Secants P N and L N intersect at point N outside of the circle. Secant P N intersects the circle at point Q and secant L N intersects the circle at point M. The length of P N is 32, the length of Q N is x, the length of L M is 22, and the length of M N is 14.

In the diagram, the length of the external portion of the secant segment PN is <u>X</u>

The length of the entire secant segment LN is <u>36</u>.

The value of x is <u>15.74</u>

Step-by-step explanation:

Snap

Jona_Fl16

7 0
3 years ago
Read 2 more answers
Which point lies on the graph of the function shown below y=-x^2+5x-3
Andrei [34K]

Answer:

C: (2,3)

Step-by-step explanation:

to solve this algebraically just plug in the x and y values of the points and see if they are true, but here is the graph to juse show you

4 0
4 years ago
James wants to compete in the international speed texting competition next year where participants compete on text speed and acc
lions [1.4K]

Complete question :

James wants to compete in the international speed texting competition next year where participants compete on text speed and accuracy. James’s current text speed is 2 characters per second. James has found that his texting speed increases at a rate of 1⁄2 a character per second for each month that he practices.

1. a. What is James’s new texting speed if he practices for only 1 month? __________

b. What is James’s new texting speed if he practices for 2 months? ____________

c. What is his new texting speed if he practices for 3 months? _____________

d. Write an algebraic equation that gives James’s texting speed s for m months of

practice.

Answer:

2 1/2 ; 3, 3 1/2 ; s = 2 + 1/2m

Step-by-step explanation:

Given that :

Speed = 2 characters per second

Rate of change = 1/2 character per second increase per month

New texting speed after 1 month :

Initial speed + (change per month * number of months)

2 + 1/2 * 1

= 2 1/2 or 2.5

After 2 months :

Initial speed + (change per month * number of months)

2 + (1/2 * 2)

2 + 1

= 3 characters per second :

After 3 months :

Initial speed + (change per month * number of months)

2 + 1/2 * 3

= 2 + 1.5

= 3.5

For the algebraic equation :

New texting speed, s

Number of months, m

Initial speed + (change per month * number of months)

s = 2 + 1/2 * m

s = 2 + 1/2m

4 0
3 years ago
Please help and dont just say hope you get it​
Llana [10]

Answer:

X = 10 Y = 14

Step-by-step explanation:

Okay, so. If you divide the Y and X for example 2.8 divided by 2 is 1.4 5.6 divided by 4 is 1.4. So like it says it is proportional. So, when you get down to the eight. You can see they are adding the X by 2 so the X is 2. The Y would be 2.8. So, you would add 2.8 into 11.2. That would be 14.

Hope this helps!

4 0
3 years ago
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