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I am Lyosha [343]
3 years ago
11

Find f(t – 3) for f(x) = 4x^2 – 8x + 4.

Mathematics
1 answer:
Vladimir79 [104]3 years ago
3 0

Answer:

<h2>A. 4t² - 32t + 64</h2>

Step-by-step explanation:

Instead of x put (t - 3) in the equation of the function f(x) = 4x² - 8x + 4:

f(t - 3) = 4(t - 3)² - 8(t - 3) + 4

<em>use (a - b)² = a² - 2ab + b² and the distributive property a(b + c) = ab + ac</em>

f(t - 3) = 4(t² - (2)(t)(3) + 3²) + (-8)(t) + (-8)(-3) + 4

f(t - 3) = 4(t² - 6t + 9) - 8t + 24 + 4

f(t - 3) = (4)(t²) + (4)(-6t) + (4)(9) - 8t + 28

f(t - 3) = 4t² - 24t + 36 - 8t + 28

f(t - 3) = 4t² + (-24t - 8t) + (36 + 28)

f(t - 3) = 4t² - 32t + 64

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expeople1 [14]

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1st. Common Difference of 3

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How would i solve a problem like this?
SVETLANKA909090 [29]

The number of combinations of size k that you can make with n items is given by the so-called binomial coefficient,

\dbinom nk = \dfrac{n!}{k!(n-k)!}

• n! is the number of ways of permuting n items.

• (n-k)! is the number of ways of permuting all but k of the n items.

Dividing n! by (n-k)! then gives the number of ways of permuting only k of the total n items.

• k! is the number of ways of permuting k items.

Dividing \frac{n!}{(n-k)!} by k! then removes all those permutations which contain the same items. We call these combinations.

For this problem we only care about counting combinations.

There are

\dbinom 63 = \dfrac{6!}{3!(6-3)!} = 20

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There are

\dbinom 72 = \dfrac{7!}{2!(7-2)!} = 21

was of selecting any 2 boys from the total 7 boys.

Then there are

\dbinom 63 \dbinom 72 = 20\cdot21 = \boxed{420}

ways of choosing a committee of 5 people consisting of 3 girls and 2 boys.

If the next question were, "What is the probability that a committee of 5 randomly selected people consists of 3 girls and 2 boys?", then you would additionally need to compute the number of ways one can make a committee of 5 people from the total 13, which is

\dbinom{13}5 = \dfrac{13!}{5!(13-5)!} = 1287

Then the probability of selecting such a committee at random is

\dfrac{\binom 63 \binom72}{\binom{13}5} = \dfrac{420}{1287} = \dfrac{140}{429} \approx 0.3263

8 0
2 years ago
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