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Leni [432]
3 years ago
11

Need help with this question

Mathematics
1 answer:
Naddik [55]3 years ago
4 0
Easiest way to graph the line is to connect the intercepts.

-50x - 3y = -1500
-3y = 50x - 1500
y = (-50/3)x + 500
y intercept (0, 500)

-50x - 3(0) = -1500
-50x = -1500
x = 300
(300, 0)

make a line that goes through the points (0, 500) and (300, 0)
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Dianne's profit from selling magazine subscriptions varies directly with the number of subscriptions sold. Dianne made $30 from
mixas84 [53]
B) I say this bcz I divided 30 by 24 and got 1.25
Profit=1.25(subscription) 
(If someone else proves this wrong then I am sorry but I am in Middle School and my teacher just taught me this and I got an 100 so if it doesn't help I'm sorry) (I even used all those math calculator and word problem websites)
4 0
3 years ago
For the function F(x)=1/x,which of these could be a value of F(x) when x is close to zero
fomenos

Answer:

C. -10,000

Step-by-step explanation:

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Answer: C. -10,000

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3 years ago
Using the graphing function on your calculator, find the solution to the system
inessss [21]
This is all I got sorry if it don’t help

4 0
2 years ago
Read 2 more answers
If a bicyclist traveled at 8:am to 8:18am a distance of 4.5 miles and from 8:18am to 8:45am a distance of 7.5 miles .What is the
Vinil7 [7]

8 am at 0

8:18 at 4.5

8:48 at 7.5 miles ???

18 min = .3 hour

48 min = .8 hour

in first .3 hours speed = 4.5/.3 = 15 mph

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3 0
3 years ago
THIS IS DUE TODAY AND I NEED AS MUCH HELP AS I CAN ITS DUE WITHIN 15 MINS
xenn [34]

Answer:

1) -3/3

2) 1/2

3) 3/4

4) -2/1

5) 1/2

6) 4/1

Step-by-step explanation:

<u>RISE OVER RUN</u>

(Rise) Using the farthest dot on the left you count how much you rise till you align with the second dot <em>(depending on whether you go up or down determines + or -)</em>

(Run) Once aligned with the second dot count whilst 'running' towards it till you've reached it.<em> (If you go right it's +, if you go left it's -)</em>

Hard to explain, but it's really easy if you look up a tutorial.

7 0
3 years ago
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