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aleksley [76]
3 years ago
13

The sum of the cost of a veterinarian exam v and a $50 x-ray tip divided equally between 2 people.

Mathematics
1 answer:
JulijaS [17]3 years ago
8 0
If you're just asking for how that'd look in algebraic form then the answer would be v+50/2
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Someone answer this for me
Andru [333]

Answer:

2 3/4

Step-by-step explanation:

I think is the right answer hope it helps

3 0
3 years ago
two pounds of grapes cost $6 . complete the table showing the price of different amounts of grapes at this rate
Degger [83]

Answer:

Table shown below

Step-by-step explanation:

To solve this problem let's use proportions.

If 2 pounds of grapes cost $6, half the amount will cost half the dollars, so the last row will have $3 in the price

For the second row, we know the price is $1, that is, one-sixth of the original given price. It should correspond to one-sixth of the amount of grapes or 2/6 pounds.

Simplifying the fraction, we get 1/3 or 0.33 pounds

6 0
3 years ago
HELP FAST, ILL GIVE BRAINIEST!!!!!
Lapatulllka [165]
The table is proportional, cause there all plus four.
-2 + 4 = 2
0 + 4 = 4
2 + 4 = 6
4 + 4 = 8

Hope it helps
3 0
3 years ago
Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the p
PIT_PIT [208]

Answer:

Colin has <em>8 sheets </em>left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers = x

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5 ...... (1)

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3 ...... (2)

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8

So, the answer is:

Colin has <em>8</em> <em>sheets </em>left for his third class.

5 0
3 years ago
⚠️HELP in boolean please⚠️
Readme [11.4K]

Answer:

how did you get those warning signs in your question

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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