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Alex
3 years ago
6

Determine if each relation is a function. Explain your reasoning.

Mathematics
1 answer:
cupoosta [38]3 years ago
5 0

Answer:

The first picture is a function while the second one is not.

Step-by-step explanation:

Reason - first picture passed the vertical line test, the other didn't.

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(04.02 LC)
True [87]
The graph is the horizontal line:

y=3  it has a slope of zero.

Function 2 is

y=3x+1, it is in slope-intercept form, y=mx+b where m is the slope and b is the y=intercept so in this case it has a slope of 3.

So the difference in slopes is 3-0=3
3 0
3 years ago
Express in simplest radical form.<br> V112<br> Answer:<br> Submit Answer
soldi70 [24.7K]
It would be 4 √7 that’s what i got
4 0
3 years ago
Need some help with this please!!
beks73 [17]

Answer:

B,C and B,A

Step-by-step explanation:

Line the cordients and go up right,Then left to find your answer.

4 0
3 years ago
Edger Anderson earns $300 A week plus a 15 percent commission only on sales he makes after his first $1000 In sales.if Mr.Anders
Mama L [17]
525 is what i got as my answer good luck





4 0
3 years ago
Read 2 more answers
5. Katie makes and sells scarves. Her monthly profit is given by P(s) = -s2 + 25s - 100, where "s" is the selling price. For wha
Marat540 [252]

Answer: s > 5

Step-by-step explanation:

-s² + 25s - 100 > 0

Coefficient of s² is -1, multiply the equation through by -1.

-1 × (-s² + 25s - 100)

s² — 25s + 100

ax² + bx + c

Then you get the factors x and y that gives x + y = b and xy = c

b = -25 and c = 100, x = -20 and y = -5

-20 × -5 = 100 and -20 + -5 = -25

Then

s² — 20s — 5s + 100 > 0

Factorising,

s (s — 20) — 5(s — 20) > 0

(s — 5)(s — 20) > 0

(s — 5) > 0 and (s — 20) > 0

s>5 and s>20

s > 5

Hope this Helps?

4 0
4 years ago
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