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2 years ago

7

Write an equation of the line passing through the points (4,8) and (-1,-12).

1 answer:

Free_Kalibri [48]2 years ago

5 0

The perpendicular would be -1/4

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PLEASE PLEASE HELP

nekit [7.7K]

Answer:68.3 degrees

Step-by-step explanation:

The diagram of the triangle ABC is shown in the attached photo. We would determine the length of side AB. It is equal to a. We would apply the cosine rule which is expressed as follows

c^2 = a^2 + b^2 - 2abCos C

Looking at the triangle,

b = 75 miles

a = 80 miles.

Angle ACB = 180 - 42 = 138 degrees. Therefore

c^2 = 80^2 + 75^2 - 2 × 80 × 75Cos 138

c^2 = 6400 + 5625 - 12000Cos 138

c^2 = 6400 + 5625 - 12000 × -0.7431

c^2 = 12025 + 8917.2

c = √20942.2 = 144.7

To determine A, we will apply sine rule

a/SinA = b/SinB = c/SinC. Therefore,

80/SinA = 144.7/Sin 138

80Sin 138 = 144.7 SinA

SinA = 53.528/144.7 = 0.3699

A = 21.7 degrees

Therefore, theta = 90 - 21.7

= 68.3 degees

8 0

3 years ago

What is the distance from point Yto wx in the figure below?

vovikov84 [41]

Answer:

C

Step-by-step explanation:

The distance YZ is sqrt(34^2-30^2)=16

6 0

3 years ago

Read 2 more answers

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

3 years ago

Can someone explain the answer to me?

Paladinen [302]

Answer:

I believe its 42

Step-by-step explanation:

i did n =14 i think and then multiplied it by 3 but i really dont exactly know

7 0

3 years ago

Name the property of real numbers illustrated by the equation (x+8)+4=x+(8+4)

FromTheMoon [43]

I think it's commutative property of addition (you can change order of addition but in the end it would be the same)

8 0

4 years ago

Read 2 more answers