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svet-max [94.6K]

2 years ago

7

I know this question is easy, but I don’t really know how to do it? Help? If possible put a quick explanation, it’s fine if not!

!

1 answer:

lorasvet [3.4K]2 years ago

3 0

Answer:

She left $10.62 the total was $58.87

Hope this helps :)

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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

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

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

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

$cot\ \varnothing = 7$

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

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

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:

$r^2 = x^2 + y^2$

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

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

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

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

Rationalize:

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

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

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

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

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

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

Rationalize

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

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

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

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

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

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

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

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

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

$cot\ \varnothing = 7$

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

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

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

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

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

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

3 0

2 years ago

PLEASE HELP EASY WORK GIVING BRAINLIEST REPORTING FALSE ANSWERS

LuckyWell [14K]

It is 70 because common sense people

7 0

2 years ago

What level of measurement is a person's "favorite sport"? A. Interval B. Nominal C. Ratio D. Ordinal

tiny-mole [99]

Answer:

(B), Nominal.

Step-by-step explanation:

A nominal or categorical scale of measurement that is generally used to label or categorize the variables.

The person's "favorite sport" depicts the nominal scale of measurement.

Hence, the correct option is (B), Nominal.

4 0

2 years ago

Solve 63 minus X equals 62 for x

r-ruslan [8.4K]

Answer:

x=1

Step-by-step explanation:

This stuff is quite easy in reality. Here's a little example including your numbers.

$63-x=62; solve for x\\\\63-x-63=62-63\\\\-x=1\\$

by this point we are left with simplifying it. Since it's a double negative we get a positive. Thus x=1

5 0

3 years ago

Using the ratio of perfect squares method, what is square root of 22 rounded to the nearest hundredth?

Alja [10]

$4+ \frac{22-16}{25-16} = \\\\ 4 + \frac{6}{9} =\\\\4 + \frac{2}{3} 
=\\\\ \boxed{\bf{4.67}}$
Using the ratio of perfect squares method, the square root of 22 rounded to the nearest hundredth is 4.67

Hope this helps!! If you have any other questions or would like further explanation just let me know!! :)

5 0

3 years ago