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2 years ago

10

Find the point-slope equation for the line that passes through the points (3,27) and (-8,-61). Use the first point in your equat

ion.

1 answer:

Llana [10]2 years ago

5 0

Answer:

$\huge\boxed{y-27=8(x-3)}$

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

where

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (3, 27) and (-8, -61).

Calculate the slope <em>m</em>:

$m=\dfrac{-61-27}{-8-3}=\dfrac{-88}{-11}=8$

Susbtitute to the equation of a line:

$y-27=8(x-3)$

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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

What is the value of -12- (-35)?

maksim [4K]

Answer:

-12-(-35)

=-12+35

=23

5 0

3 years ago

Read 2 more answers

There are 2,560 fl oz. in a fish tank, how many gallons is the fish tank? help me !!!!!

vfiekz [6]

Answer:

Your answer would be 12

Step-by-step explanation:

Hopefully this helped, if not HMU and I will try my best to get you a better answer! :)

5 0

3 years ago

The roof of one building is 400 feet below the roof of another building. A person standing on the lower roof sees another person

m_a_m_a [10]

The distance between the two buildings is 844.23 feet apart, we go this answer by simply applying the SOH CAH TOA method

<h3>Trigonometry</h3>

Given Data

Height of Building A (Opp) = 400 feet

Angle of Elevation θ = 35°

Distance between the buildings Adj = ??

Applying the SOH CAH TOA Approach we have

Tan θ = Opp/Adj

Substituting our given data we have

Tan 35 = 400/x

0.4738 = 400/x

make x subject of the formula we have

x = 400/0.4738

x = 844.23 feet

learn more about trigonometry here:

brainly.com/question/24349828

7 0

2 years ago

The length of a rectangle is 6 units longer that the width , w.

Ulleksa [173]

No B

lets say W is 5

C is saying 4x5+6 right?

so all 4 side are 5 and plus a random six

W is only equal to 2 sides so C and D are gone

A. 2 x 5 + 6 is saying
2 sides are 5 + 6 and then a missing side

but B is 2 x 5 + 12
B could also say 2 x 5 + 2 x 6
2 sides are 5 and then 6 side are 5

Your answer is B

3 0

3 years ago