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Montano1993 [528]
2 years ago
9

Helpppppp me please

Mathematics
1 answer:
nataly862011 [7]2 years ago
7 0

Answer:Sure thing? But first what do you need help with and maybe just maybe I will help you with whatever you need help with?

Step-by-step explanation:

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Under his cell phone plan, Liam pays a flat cost of $69.50 per month and $5 per gigabyte. He wants to keep his bill under $90 pe
natulia [17]

Answer:

5y+69.50<90

Step-by-step explanation:

You can think of the number of gigabytes he will use as "y", since we don't know the number of gigabytes he will use. The flat cost of one month is $69.50.

You need to add those to calculate the complete cost. The inequality would use a less than sign because he needs to pay less than $90/month. It would not use an inequality sign with a line under it because the problem doesn't say he wants to use $90 or less.

You would solve using basic computation (subtraction and division). Hope this helped!

3 0
2 years ago
The list below shows the results of a survey of favorite ice-cream flavors among middle-school students.
Mrac [35]

Answer:

40%

Step-by-step explanation:

18+12+20=50

20/50×100=

40%

8 0
3 years ago
Help pls ! i need help please
Nuetrik [128]

Answer:

115 degrees

Step-by-step explanation:

The line RQ meets at 115 degrees (Look at the bottom numbers)

It's not 65 degrees because the angle is obtuse.

6 0
3 years ago
A polynomial function has a root of –6 with multiplicity 1, a root of –2 with multiplicity 3, a root of 0 with multiplicity 2, a
densk [106]
The correct answer for the question that is being presented above is this one: "C.) The graph of the function is positive on (–2, 4)." A polynomial function has a root of –6 with multiplicity 1, a root of –2 with multiplicity 3, a root of 0 with multiplicity 2, and a root of 4 with multiplicity 3. If the function has a positive leading coefficient and is of odd degree, then <span>C.) The graph of the function is positive on (–2, 4).</span>
6 0
3 years ago
Read 2 more answers
P(x)= 3x^5+2x^4+x^3-10x-8
erik [133]
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3 0
3 years ago
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