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makkiz [27]
3 years ago
15

food ,treats ,toys ,bed, collar ,leash, Veterinary Care ,daily dog walking ,boarding ,and training classes which ones go under v

ariable expenses and fixed expenses?
Mathematics
2 answers:
Aneli [31]3 years ago
4 0
Variable Expenses: Food, Treats, Toys, Bed, Collar, Leash
Fixed Expenses: Veterinary Care, Daily Dog Walking, Boarding, Training Classes
Salsk061 [2.6K]3 years ago
3 0
They would be variable expenses because you don't pay for all of them every period of time.
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I'd really appreciate it if anyone helped! :) Giving Brainliest!
Anna11 [10]

the first three are irrational


5 0
2 years ago
5 less the quotient of 14 times a number x and 9
AURORKA [14]

Answer:

\frac{14x}{9}-5

Step-by-step explanation:

14 times a number x :  14* x = 14x

Quotient of 14 times a number x  an 9 :   \frac{14x}{9}

5 less the quotient of 14 times a number x and 9 :

\frac{14x}{9}-5

7 0
2 years ago
I really don’t understand this question and I don’t know what to do
trasher [3.6K]
I would assume it is 0% for the minimum as there is a chance that rangers didn’t see anything and 65% for the highest as that is all the people that saw giraffe. I am really unsure though.

My second answer is minimum is 55.25% and the maximum is 65%. But once again I ain’t sure and logically I would go for my first answer
5 0
3 years ago
Aleena expands 2xx+3-5(x-2) as 2x2+3-5x-10. Has she expanded it correctly? Please identify the mistakes and write the correct wa
Nimfa-mama [501]

The given expression is :

2xx+3-5(x-2)

We know that, x{\cdot}x=x^2

=2x^2+3-5(x-2)\\\\=2x^2+3+(-5)(x)+(-5)(-2)\\\\=2x^2+3-5x+10

Aleena expands 2xx+3-5(x-2) as 2x²+3-5x-10. She is mistaken because the sign before 10 should be +10 instead of -10. Hence, the correct expanded form is 2x²+3-5x+10. Hence, this is the required solution.

5 0
2 years ago
A freight train leaves a station and travels north at 60 mph. 4 hours ​later, a passenger train leaves on a parallel track and t
harkovskaia [24]

Answer:

first do:

60mph * 4 hours= 240mi

Let D = the distance the passenger train has to  

travel to catch the freight train

Start a stopwatch when the passenger train leaves

Let t = the time on the stopwatch when they meet

Equation for freight train:

d-240=60t

Equation for passenger train:

d=100t

all together:

100t-240=60t

we can do 100t-60t because they both have (t)= 40t

40t=240 *now divide 240 by 40= 6

THEN do:

D= 100mph x 6= 600mi

They will meet 600 mi from the station

to check:

d-240=60t x 6

600-240= 360

60 x 6= 360

360=360

5 0
3 years ago
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