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Lina20 [59]
2 years ago
11

If (x – 3)(x + 5) = 49, then what is the value of the expression (x – 4)(x + 6)?

Mathematics
1 answer:
mina [271]2 years ago
7 0
It’s impossible to do this because…….. so that’s why it’s impossible
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0.04 rounded to the nearest hundreth<br><br> and <br><br> 0.2% rounded to the nearest hundreth
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To round a number, you look to the next place to the right of what you want to round to... for example, if you want to round to the nearest hundredth, you look to the thousandths place to see whether the hundredths place rounds up or down.

0.04 is already at the hundredths place so the thousandths place is zero... the answer to the nearest hundredth is 0.04.

For 0.2%, you have to convert the percentage to a decimal by dividing by 100 (move the decimal 2 places to the left).

0.2% = 0.002

So now to round to the nearest hundredth, we look to the thousandths place. The thousandths place has a 2 in it so the hundredths place rounds down. 0.2% total he nearest hundredth is 0.
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Simplify the following expression: 3m-5n+6m-8n
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3m - 5n + 6m - 8n \\ 3m + 6m - 5n - 8n \\ 9 m- 13n

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A(b + c) = a*(b + c) = a*b + a*c

You must multiply individual terms and see what it would equal
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Tommy is a video game designer at a new start-up company. How much does he make per month?
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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
3 years ago
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