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3 years ago

Solve the following equation for 0°≤x≤360° : 2 cot 2x + 3 = 0

Mathematics

1 answer:

evablogger [386]3 years ago

3 0

Answer:

⋅

Explanation:

Factorizing your equation we get

cot

(

)

cos

(

)

we get the following Solutions in the given interval

for the second case we get

cot

(

)

−

cos

(

)

−

sin

(

)

and we get

Hint:

To get the Solutions of the last equation you can use

√

sin

(

)

Step-by-step explanation:

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10-x=-2x what is it ???

aniked [119]

Answer:

x = - 10

Step-by-step explanation:

Given

10 - x = - 2x ( add 2x to both sides )

10 + x = 0 ( subtract 10 from both sides )

x = - 10

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3 years ago

What is the sum of 15 and the opposite of 13?

NemiM [27]

Answer:

Step-by-step explanation:

15+-13 = 2

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4 years ago

One point on the trend line is 200,6 using this point and the y-intercept find the slope of the trend line.

Marina CMI [18]

Step-by-step explanation:

Use the formula (Y2 - Y1)/(X2 - X1) to find the slope between two points

We'll make Point 1 (which is X1 and Y1) the Y-intercept so

X1, Y1 = (0, 5.00)

And we'll make Point 2 (which is X2 and Y2) the point on the trend line

X2, Y2 = (200, 6.00)

Plug into the formula:

(6.00 - 5.00)/(200 - 0)

= 1/200 or 0.005

Slope: 1/200 or 0.005

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6 0

3 years ago

Use the Pythagorean theorem and the following diagram to help you find the area and perimeter of the following triangle. Please

Mekhanik [1.2K]

Answer:

Perimeter = 30

Area = 30

Step-by-step explanation:

$(x+8)^2 -x^2 = 12^2$

$x^2 +16x +64 - x^2 = 144$

$16x+64=144$

$16x = 80$

$x = 5$

Double check:

$\sqrt{12^2 + 5^2} = (5+8)\\\sqrt{12^2 + 5^2} = 13\\13 = 13$

Perimeter:

$12+5+13=30$

Area( $\frac{1}{2}bh$ ):

$\frac{1}{2}$ × 12 × 5 = 30

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3 years ago

Read 2 more answers

Using the defects method, which of these relationships represents the Law of Cosines if the measure of the included angle betwee

horrorfan [7]

Asked and answered elsewhere.
brainly.com/question/1410592

_____
Google doesn't recognize the terms "defects method" except in association with various postings of this same problem.

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4 years ago