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1 year ago

7

Evaluate cos 25o to four decimal places.

1 answer:

Firdavs [7]1 year ago

8 0

$\tt \cos(25) = 0.991203 \: or \: 0.99$

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Which equation is y = –6x^2 + 3x + 2 rewritten in vertex form? y = negative 6 (x minus 1) squared + 8 y = negative 6 (x + one-fo

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Answer:

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

Step-by-step explanation:

Given:

$y = -6x^2 + 3x + 2$

Required

Rewrite in vertex form

The vertex form of an equation is in form of: $y = a(x - h)^2+ k$

Solving: $y = -6x^2 + 3x + 2$

Subtract 2 from both sides

$y - 2 = -6x^2 + 3x + 2 - 2$

$y - 2 = -6x^2 + 3x$

Factorize expression on the right hand side by dividing through by the coefficient of x²

$y - 2 = -6(x^2 + \frac{3x}{-6})$

$y - 2 = -6(x^2 - \frac{3x}{6})$

$y - 2 = -6(x^2 - \frac{x}{2})$

Get a perfect square of coefficient of x; then add to both sides

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

<em>Rough work</em>

The coefficient of x is $\frac{-1}{2}$

It's square is $(\frac{-1}{2})^2 = \frac{1}{4}$

Adding inside the bracket of $-6(x^2 - \frac{x}{2})$ to give: $-6(x^2 - \frac{x}{2} + \frac{1}{4})$

To balance the equation, the same expression must be added to the other side of the equation;

Equivalent expression is: $-6(\frac{1}{4})$

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

The expression becomes

$y - 2 -6(\frac{1}{4})= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{6}{4}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{3}{2}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

Factorize the expression on the right hand side

$y - 2 -\frac{3}{2}= -6(x - \frac{1}{2})^2$

$y - (2 +\frac{3}{2})= -6(x - \frac{1}{2})^2$

$y - (\frac{4+3}{2})= -6(x - \frac{1}{2})^2$

$y - (\frac{7}{2})= -6(x - \frac{1}{2})^2$

$y +\frac{7}{2} = -6(x - \frac{1}{2})^2$

Make y the subject of formula

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

<em>Solved</em>

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