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2 years ago

Rewrite the function by completing the square. f(x)=x^2+16x-46

Mathematics

1 answer:

andrew-mc [135]2 years ago

8 0

Answer:

$f(x)=(x+8)^{2}-110$

Step-by-step explanation:

Completing the square is a multi-step process that involved rewriting an equation such that it has a factored term, a coefficient, and a constant. First, group the equation, then add a term such that the grouped part of the equation can be simplified, then balance the equation, finally simplify and factor the equation.

$f(x)=x^{2} +16x - 46\\$

Group;

$f(x)=(x^{2}+16)-46$

Add in a term that makes the grouped part of the equation a perfect square, this can be done by dividing the linear term by 2, then balancing the equation;

$f(x)=(x^{2}+16x+64)-46-64$

Simplify, factor;

$f(x)=(x+8)^{2} - 110$

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3 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

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3 years ago

in a rectangle, the ratio of the width to the length is 4:5. If the rectangle is 40 centimeters long find its width

dybincka [34]

Gofipanswer.com copy and paste ur answer it showes the answer and explains it good luck god bless<3

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3 years ago

21 is 84% of what number?

Radda [10]

21 is 84% of 25

Convert the percentage (84) into a decimal by dividing it over 100:
$\frac{84}{100} = 0.84$

Divide 21 over 0.84:
$\frac{21}{0.84} = 25$

7 0

3 years ago

Read 2 more answers

Rewrite the function by completing the square. f(x)=x^2+16x-46 ​

Rewrite the function by completing the square. f(x)=x^2+16x-46