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Orlov [11]
2 years ago
8

Rewrite the function by completing the square.

Mathematics
1 answer:
ddd [48]2 years ago
6 0

Answer:

  f(x) = (x -6)² +14

Step-by-step explanation:

Completing the square involves writing part of the function as a perfect square trinomial.

<h3>Perfect square trinomial</h3>

The square of a binomial results in a perfect square trinomial:

  (x -h)² = x² -2hx +h²

The constant term (h²) in this trinomial is the square of half the coefficient of the linear term: h² = ((-2h)/2)².

<h3>Completing the square</h3>

One way to "complete the square" is to add and subtract the constant necessary to make a perfect square trinomial from the variable terms.

Here, we recognize the coefficient of the linear term is -12, so the necessary constant is (-12/2)² = 36. Adding and subtracting this, we have ...

  f(x) = x² -12x +36 +50 -36

Rearranging into the desired form, this is ...

  f(x) = (x -6)² +14

__

<em>Additional comment</em>

Another way to achieve the same effect is to split the given constant into two parts, one of which is the constant necessary to complete the square.

  f(x) = x² -12x +(36 +14)

  f(x) = (x² -12x +36) +14

  f(x) = (x -6)² +14

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Compute delta y/delta x for the interval [2,5], where y = 4x - 9.
Kamila [148]
We have to compute: Δ y / Δ x.
The interval is [ 2, 5 ] and y = 4 x - 9.
y 1 = 4 · 2 - 9 = 8 - 9 = - 1
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3 0
3 years ago
Solve the system by substitution.<br> y = -9x<br> y = 2x + 33
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5 0
3 years ago
A box contains 24 transistors,4 of which are defective. If 4 are sold at random,find the following probabilities. i. Exactly 2 a
zavuch27 [327]

SOLUTION

This is a binomial probability. For i, we will apply the Binomial probability formula

i. Exactly 2 are defective

Using the formula, we have

\begin{gathered} P_x=^nC_x\left(p^x\right?\left(q^{n-x}\right) \\ Where\text{ } \\ P_x=binomial\text{ probability} \\ x=number\text{ of times for a specific outcome with n trials =2} \\ p=\text{ probability of success = }\frac{4}{24}=\frac{1}{6} \\ q=probability\text{ of failure =1-}\frac{1}{6}=\frac{5}{6} \\ ^nC_x=\text{ number of combinations = }^4C_2 \\ n=\text{ number of trials = 4} \end{gathered}

Note that I made the probability of being defective as the probability of success = p

and probability of none defective as probability of failure = q

Exactly 2 are defective becomes the binomial probability

\begin{gathered} P_x=^4C_2\times\lparen\frac{1}{6})^2\times\lparen\frac{5}{6})^{4-2} \\ P_x=6\times\frac{1}{36}\times\frac{25}{36} \\ P_x=\frac{25}{216} \\ =0.1157 \end{gathered}

Hence the answer is 0.1157

(ii) None is defective becomes

\begin{gathered} \lparen\frac{5}{6})^4=\frac{625}{1296} \\ =0.4823 \end{gathered}

hence the answer is 0.4823

(iii) All are defective

\begin{gathered} \lparen\frac{1}{6})^4=\frac{1}{1296} \\ =0.00077 \end{gathered}

(iv) At least one is defective

This is 1 - probability that none is defective

\begin{gathered} 1-\lparen\frac{5}{6})^4 \\ =1-\frac{625}{1296} \\ =\frac{671}{1296} \\ =0.5177 \end{gathered}

Hence the answer is 0.5177

3 0
1 year ago
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