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3 years ago

14

Find the zeros for (x+7) (x+5)+0

1 answer:

JulijaS [17]3 years ago

3 0

Steps to solve for the zero:

(x + 7)(x + 5) = 0

~Set both factors to equal zero

x + 7 = 0

x + 5 = 0

~Solve for x in [ x + 7 = 0 ]

x + 7 = 0

x + 7 - 7 = 0 - 7

x = -7

~Solve for x in [ x + 5 = 0 ]

x + 5 = 0

x + 5 - 5 = 0 - 5

x = -5

Therefore, the solutions are x = -7 and x = -5

Best of Luck!

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Step-by-step explanation:

Given

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Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

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$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

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<u>Solving the trigonometry functions</u>

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$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

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$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

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Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

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Read 2 more answers