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2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!

Mathematics

1 answer:

ioda2 years ago

4 0

Answer: Vertex = (0, 3) 2nd point = (1, 2)

<u>Step-by-step explanation:</u>

f(x) = -x² + 3 → f(x) = a(x - 0)² + 3

If you use transformations with the parent graph f(x) = x², the transformation is reflected over the x-axis and shifted 3 units up.

If you use the vertex formula: f(x) = a(x - h)² + k

the vertex is (0, 3) ← PLOT THIS COORDINATE
"a" is negative, so the parabola points downward

This equation does not factor over the rational numbers, so find a second point by choosing an x-value and plugging it into the given equation to solve for y. <em>I chose x = 1</em>

f(1) = -(1)² + 3

= -1 + 3

= 2

So, an additional point is (1, 2) ← PLOT THIS COORDINATE

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The length of a rectangle is 5 inches less than 4 times the width. The perimeter is 110 inches. Find the length and the width.

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Step-by-step explanation:

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Answer:

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

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

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

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

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

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

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

Where:

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$r = \sqrt{x^2 + y^2$

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So:

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<u>Solving the trigonometry functions</u>

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

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

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

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

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

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

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