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2 years ago

Use a graphing calculator to solve the equation -3 cost= 1 in the interval from . Round to the nearest hundredth.

Mathematics

1 answer:

RSB [31]2 years ago

8 0

In place of t, or theta, I'm going to utilize x instead. So the equation is -3*cos(x) = 1. Get everything to one side and we have -3*cos(x)-1 = 0

Let f(x) = -3*cos(x)-1. The goal is to find the root of f(x) in the interval [0, 2pi]

I'm using the program GeoGebra to get the task done of finding the roots. In this case, there are 2 roots and they are marked by the points A and B in the attachment shown

A = (1.91, 0)
B = (4.37, 0)

So the two solutions for theta are
theta = 1.91 radians
theta = 4.37 radians

You might be interested in

Segment FG begins at point F(-2, 4) and ends at point G(-2, -3). The segment is translated 3 units left and 2 units up, then ref

Alexus [3.1K]

Answer:

$F'G' = 7$

Step-by-step explanation:

Given

$F = (-2,4)$

$G = (-2,-3)$

Required

Distance of F'G'

The transformation that give rise to F'G' from FG are:

Translation
Reflection

The above transformations are referred to as rigid transformation, and as such the side lengths remain unchanged.

i.e.

$F'G' = FG$

Calculating FG, we have:

$FG = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

Where:

$F = (-2,4)$ --- $(x_1,y_1)$

$G = (-2,-3)$ --- $(x_2,y_2)$

$FG = \sqrt{(-2 - -2)^2 + (4 - -3)^2}$

$FG = \sqrt{(0)^2 + (7)^2}$

$FG = \sqrt{0 + 49}$

$FG = \sqrt{49}$

Take positive square root

$FG = 7$

Recall that:

$F'G' = FG$

$F'G' = 7$

6 0

2 years ago

Write the equation of the line that passes through the point (-3, 6) and is parallel to the line

Ugo [173]

Answer:

$\because \:l_{1} \parallel \: l_{2} \\ \therefore m_{1} = m_{2} = \frac{1}{2} \\ \frac{y - 6}{x + 3} = \frac{1}{2} \\ y - 6 = \frac{1}{2}x + \frac{3}{2} \\ y = \frac{1}{2}x + \frac{3}{2} + 6\\ y = \frac{1}{2}x + \frac{15}{2}$