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expeople1 [14]
3 years ago
13

nick is given 450 to sepend on a vacatio. he decides to spend $5 a day the amount nick has left and the number of days are relat

ed
Mathematics
1 answer:
omeli [17]3 years ago
7 0

Answer:

m= 450 - (5*n)

Step-by-step explanation:

m is the money left

n is the days

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PLZ I NEED THIS ASAP!! ILL GIVE YOU BRAINLIST!!!
Aloiza [94]

Answer:

Yes

Step-by-step explanation:

I feel like its yes because x and y share the numbers 1-4

4 0
2 years ago
Read 2 more answers
What's the answer to this
lyudmila [28]
8(14-9)+5
________
2
3. + 6


8(5)+5
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9 + 6


40 + 5
————
15


45
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= 3
4 0
3 years ago
ONLY ANSWER THIS IF YOU HAVE AN EXPLANATION AND IF YOU KNOW WHAT THE ANSWER IS.
Bad White [126]
Pretty sure it’s G cause everyone does math problems at a different pace. And also some math problems are more work compared to others ig. so G would have the most variability for the data
6 0
2 years ago
Solve: |11x + 10| = 22
Tju [1.3M]

Answer:

x = 12/11, -32/11

Step-by-step explanation:

4 0
2 years ago
Question 5 and 6 please help me
sp2606 [1]

Problem 5

The function is continuous for the given domain x \ge 6

This is because y = (-5/6)x+5 is itself continuous, and any interval subset of this function is also continuous. We can plug in any real number that is equal to 6 or larger, and get some y output. If we plugged in x = 6, then we'd get

y = (-5/6)x+5

y = (-5/6)*6 + 5

y = -5+5

y = 0

This is the largest y value possible. Why? Because y = (-5/6)x+5 has a negative slope, so the graph is going downhill as you read it from left to right. As x gets bigger, y gets smaller. The smallest x value allowed in the domain produces the largest y value in the range. There is no smallest y value as the y values keep going down forever.

The range is therefore y \le 0

In interval notation, you can write the range as (-\infty, 0]. The square bracket indicates "include this endpoint as part of the range".

======================================================

Problem 6

The function is discrete for this given domain. The domain itself is a discrete list of values. We cannot plug in values between say 0 and 2. We can only substitute one of those values from the list given. Consequently, the y values will also be a list, and not an interval like problem 5 had.

-----------

If you plugged in x = -4, then you should get...

y = (-1/2)*(-4)+2

y = 2+2

y = 4

So the input x = -4 lead the output y = 4

Repeat for x = -2

y = (-1/2)x+2

y = (-1/2)*(-2)+2

y = 1+2

y = 3

and the same for x = 0 as well

y = (-1/2)x+2

y = (-1/2)*0 + 2

y = 0 + 2

y = 2

and x = 2 also

y = (-1/2)x+2

y = (-1/2)*2 + 2

y = -1+2

y = 1

Finally, plug in x = 4

y = (-1/2)x+2

y = (-1/2)*4+2

y = -2+2

y = 0

---------------

If we plugged each of these x values {-4, -2, 0, 2, 4} one at a time into the equation y = (-1/2)x+2, then we get this list of values {4, 3, 2, 1, 0}

Sorting the values from smallest to largest, we have this range {0, 1, 2, 3, 4}

3 0
2 years ago
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