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user100 [1]
1 year ago
9

Brainliest!!!!!!

Mathematics
2 answers:
Nataly [62]1 year ago
4 0

Answer:

2x² - 12x + 13

Step-by-step explanation:

Given following:

  • k(x) = 2x² - 5
  • p(x) = x - 3

Start solving from right to left across the function.

Steps:

  • k[p(x)]
  • k[x - 3]
  • 2(x - 3)² - 5
  • 2(x² - 6x + 9) - 5
  • 2x² - 12x + 18 - 5
  • 2x² - 12x + 13
SOVA2 [1]1 year ago
4 0

Answer:

2x² - 12x + 13

Step-by-step explanation:

Steps:

k[p(x)]

k[x - 3]

2(x - 3)² - 5

2(x² - 6x + 9) - 5

2x² - 12x + 18 - 5

2x² - 12x + 13             pls brainlest

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How do I solve this equation for x?
tester [92]
1) Divide both sides by 2
2x+12^2/3 =16

2) Solve the exponent
2x+12^2/3 =16

(2x+12^2/) ^3/2 =16^3/2 ( raise both sides to power)

2x+12 = 64

3) Subtract 12 from both sides
2x + 12 - 12 = 64-12

2x = 52

4) divide both sides by 2
2x/2 = 52/2

x = 26


Final answer : x = 26

7 0
2 years ago
What is 3,256,423,865 rounded to the nearest million?
serg [7]
3,256,423,865 rounded to the nearest million is 3,256,000,000. How I know is that the 4 in the hundred thousands place is lower than 5. So the 6 would of stayed the same and the rest of the number behind 4 would of been 0, including the 4.
5 0
3 years ago
One hundred items are simultaneously put on a life test. Suppose the lifetimes
romanna [79]

Answer:

a) \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b) \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( i-1 )st and the ith failures. Then, the T_{i} are independent with \mathrm{T}_{\mathrm{i}} being exponential with rate \frac{(101-i)}{200} .

Therefore,

a) E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]

=\sum_{i=1}^{5} \frac{200}{101-i}

\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b)

The variance is given by, \mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]

\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

7 0
3 years ago
Math question below!!​
nikdorinn [45]

Answer:

A

Step-by-step explanation:

x² - 4x - 5 = 0

x² - 5x + x - 5 = 0

x(x - 5) + (x -5) = 0

(x - 5)(x + 1) = 0

x - 5 = 0   ; x + 1 = 0

x = 5      ; x = -1

7 0
3 years ago
How is combining "like terms" similar to adding and subtracting whole numbers?
arsen [322]
Because like terms you add or subtract together to make it easier to do the problem. just like you do with whole numbers. you add an subtract to make the answer simpler.
8 0
2 years ago
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