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blsea [12.9K]
3 years ago
9

I need help completing this problem ASAP

Mathematics
1 answer:
aalyn [17]3 years ago
5 0

4/(√<em>x</em> - √(<em>x</em> - 2)) × (√<em>x</em> + √(<em>x</em> - 2))/(√<em>x</em> + √(<em>x</em> - 2))

= 4 (√<em>x</em> + √(<em>x</em> - 2)) / ((√<em>x</em>)² - (√(<em>x</em> - 2))²)

= 4 (√<em>x</em> + √(<em>x</em> - 2)) / (<em>x</em> - (<em>x</em> - 2))

= 4 (√<em>x</em> + √(<em>x</em> - 2)) / (<em>x</em> - <em>x</em> + 2)

= 4 (√<em>x</em> + √(<em>x</em> - 2)) / 2

= 2 (√<em>x</em> + √(<em>x</em> - 2))

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Step-by-step explanation:

0+1 = 1

1/2+1/2= 1

-50 + 51= 1

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Match each function with the corresponding function formula when h(x) = 2 – 2x and g(x) = –2 x + 2.
Minchanka [31]

the correct functions are

h(x) = 3x + 2

g(x) = 2^x

proceed to calculate the different values ​​of K (x)

case 1) k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + 2]-----------> is the option 4. k(x) = 2^(3x + 2 )

case 2) k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+2--------> is the option 1.) k(x) = 2^x + 3x + 4


case 3)k(x) = g(x) - h(x)

k(x) = [2^x]-[3x + 2]=2^x-3x-2--------> is the option 2.) k(x) = 2^x - 3x - 2

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P.S ( The calculator given to me doesnt work ;-; ) NEED HELP ASAP FOR MY HOMEWORK ASSIGNMENT!! IF YOU ANSWER RIGHT ILL MARK YOU
wolverine [178]

Answer for the first box:  20

Answer for the second box: 6

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

Explanation:

The notation a(1) represents the first term of the sequence given. The first term is 20, so thats why a(1) = 20.

The notation a(n) = a(n-1) + ____ means we add some term to the previous term to get the next. This process happens over and over. In this case, we add 6 to each term to get the next

  • term1 = 20
  • term2 = term1 + 6 = 20 + 6 = 26
  • term3 = term2 + 6 = 26 + 6 = 32
  • term4 = term3 + 6 = 32 + 6 = 38

and so on...

So we write a(n) = a(n-1) + 6

This value of 6 can be found by selecting any term and subtracting off the previous term, so we could say any of the following below

  • term2 - term1 = 26 - 20 = 6
  • term3 - term2 = 32 - 26 = 6
  • term4 - term3 = 38 - 26 = 6

That's how we end up with

\begin{cases}a(1) = 20\\a(n) = a(n-1) + 6\end{cases}

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

Extra info:

The common difference is d = 6 and the first term is a = 20

We can use these two facts to find the nth term a(n)

a(n) = a + d*(n-1)

a(n) = 20 + 6*(n-1)

a(n) = 20 + 6n - 6

a(n) = 6n + 14

Then note how plugging in n = 1 leads to

a(n) = 6*n + 14

a(1) = 6*1 + 14

a(1) = 20

and n = 2 leads to

a(n) = 6*n + 14

a(2) = 6*2 + 14

a(2) = 26

and so on

This allows us to plug in any value of n we want, without having to generate the previous terms before it. So for instance, you could plug in n = 100 to jump directly to the 100th term. If you use the recursive definition in the last section, then you have to generate the first 99 terms first before you can get to the 100th term.

7 0
3 years ago
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