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bija089 [108]
3 years ago
11

Sam can spend no more than $75 on school supplies. he has to spend $12 on paper and pencils. what is the maximum amount sam can

spend on a graphing calculator?
Mathematics
2 answers:
Pachacha [2.7K]3 years ago
7 0

Answer:

$63

Step-by-step explanation:

Let x represent the amount spent on graphing calculator.

We have been given given that Sam has to spend $12 on paper and pencils. The total cost of graphing calculator, paper and pencils would be x+12.

Since Sam can spend no more than $75 on school supplies, so the amount spent on school supplies should be less than or equal to 75.

We can represent this information in an inequality as:

x+12\leq 75

Let us solve for x:

x+12-12\leq 75-12

x\leq 63

Therefore, Sam can spend a maximum amount of $63 on the graphing calculator.

soldier1979 [14.2K]3 years ago
3 0
Its 63 dollars
----------------
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There are 12 squares in each 4 x 3 grid.

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This means that there should be 3 shaded squares in the grid.

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Question 7
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2 years ago
Determine the first four terms of the sequence in which the nth term is a_n=(n+1)!/(n+2)!
Vilka [71]

Answer: Choice B) 1/3, 1/4, 1/5, 1/6

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

Plug in n = 1 to find the first term

a_n = ( (n+1)! )/( (n+2)! )

a_1 = ( (1+1)! )/( (1+2)! )

a_1 = ( 2! )/( 3! )

a_1 = ( 2*1 )/( 3*2*1 )

a_1 = 2/6

a_1 = 1/3

The first term is 1/3. Optionally you can stop here because only choice B has 1/3 listed as the first term, so this must be the answer. However, I'm going to keep going to show how to find the three other terms. This will help confirm why choice B is the answer, and it will be handy for those times when you aren't given multiple choice answers.

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

Plug in n = 2

a_n = ( (n+1)! )/( (n+2)! )

a_2 = ( (2+1)! )/( (2+2)! )

a_2 = ( 3! )/( 4! )

a_2 = ( 3*2*1 )/( 4*3*2*1 )

a_2 = 6/24

a_2 = 1/4

The second term is 1/4

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

Plug in n = 3

a_n = ( (n+1)! )/( (n+2)! )

a_3 = ( (3+1)! )/( (3+2)! )

a_3 = ( 4! )/( 5! )

a_3 = ( 4*3*2*1 )/( 5*4*3*2*1 )

a_3 = 24/120

a_3 = 1/5

The third term is 1/5

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

Plug in n = 4

a_n = ( (n+1)! )/( (n+2)! )

a_4 = ( (4+1)! )/( (4+2)! )

a_4 = ( 5! )/( 6! )

a_4 = ( 5*4*3*2*1 )/( 6*5*4*3*2*1 )

a_4 = 120/720

a_4 = 1/6

The fourth term is 1/6

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

The first four terms are: 1/3, 1/4, 1/5, 1/6, so that confirms why choice B is the answer.

4 0
3 years ago
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