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ycow [4]
3 years ago
13

Write a system of linear equations for the problem below(you do not have to solve it) Twenty-six students attended the community

service event. The number of girls was 4 fewer than twice the number of boys. How many boys attended the community service event?
Mathematics
1 answer:
anzhelika [568]3 years ago
7 0

Answer:

Number of boys who attended the community service event = 10

Step-by-step explanation:

Let x denotes number of girls and y denotes number of boys.

Twenty-six students attended the community service event.

So,

x+y=26

The number of girls was 4 fewer than twice the number of boys.

x=2y-4

Put x=2y-4 in x+y=26

2y-4+y=26\\3y=26+4\\3y=30\\y=10

So,

Number of boys who attended the community service event = 10

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vichka [17]
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10 + (120 + -9x) = -9x
 
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Combine like terms: -9x + 9x = 0 130 = 0
 
Solving 130 = 0
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4 years ago
PLEASE HELP 25 POINTS!!!
miss Akunina [59]

Sequence: 2 + 1 + 1/2 + 1/4 + 1/8 + ...

Explanation:

Here Given Sum to Infinity

\sf S \infty  \ =a_1  \ x \ \dfrac{1}{1-r}

Identify following

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Solve:

\rightarrow \sf S \infty  \ =2  \ x \ \dfrac{1}{1-\frac{1}{2} }

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2 years ago
Give the following number in Base 16<br>NO LINKS!!! ​
olga nikolaevna [1]

Answer:

\large{\boxed{35_{10} = \sf 23_{16}}}

Explanation:

\sf Given : 35_{10}

Compute:

35 ÷ 16 = 2.1875 = 2R3

2 ÷ 16 = 0.125 = 0R2

Jot down the remainders:

  • 3        ↑
  • 2        ↑

======

\sf Answer :  \ 35_{10}   =  23_{16}

\hrulefill

Let's look at another example:  \sf 325_{10} = [?]_{16}

Compute:

325 ÷ 16 = 20.3125 = 20R5

20 ÷ 16 = 1.25 = 1R4

1 ÷ 16 = 0.0625 = 0R1

Jot down remainders:

  • 5        ↑
  • 4        ↑
  • 1         ↑

=====

\sf 325_{10} = [145]_{16}

\hrulefill

Another example: \sf 500_{10} = [?]_{15}

Compute:

500 ÷ 15 = 33.33... = 33R5

33 ÷ 15 = 2.2 = 2R3

2 ÷ 15 = 0.13... = 0R2

Jot down remainders:

  • 5      ↑
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=====

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7 0
2 years ago
A student solves the following equation and determines that the solution is −2. Is the student correct? Explain. 3 a + 2 − 6a a2
Ganezh [65]

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Then, we multiply both sides of the equation by (a-2) (a + 2):

\frac{3(a-2)(a+2)}{a+2} - \frac{6a(a-2)(a+2)}{(a-2)(a+2)}  = \frac{(a-2)(a+2)}{a-2}

Later, canceling similar terms we have:

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Answer:

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Answer:

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