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3 years ago

V1 - 2 sin cos 0 = cos 0 - sin

1 answer:

kramer3 years ago

4 0

$\sqrt{1 - 2sin θ \cosθ } = \sqrt{ {sin}^{2} θ + {cos}^{2}θ - 2 sinθcosθ } = \sqrt{ {(cosθ - sinθ) }^{2} } = cosθ - sinθ$

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Angle∠efg and angle∠gfh are a linear pair, mangle∠efgequals=55nplus+2020, and mangle∠gfhequals=33nplus+1616. what are mangle∠e

mylen [45]

A linear pair is a pair of angles that form a line. When added, they will equal 180.

< EFG + < GFH = 180
5n + 20 + 3n + 16 = 180
8n + 36 = 180
8n = 180 - 36
8n = 144
n = 144/8
n = 18

< EFG = 5n + 20.....5(18) + 20 = 110 <==
< GFH = 3n + 16.....3(18) + 16 = 70 <==

3 0

4 years ago

Read 2 more answers

A: What are the solutions to the quadratic equation x2+9=0? B: What is the factored form of the quadratic expression x2+9? Selec

Paraphin [41]

Answer:

A. The solutions are $x=3i,\:x=-3i$ .

B. The factored form of the quadratic expression $x^2+9=(x-3i)(x+3i)$

Step-by-step explanation:

A. To find the solutions to the quadratic equation $x^2+9=0$ you must:

$\mathrm{Subtract\:}9\mathrm{\:from\:both\:sides}\\\\x^2+9-9=0-9\\\\\mathrm{Simplify}\\\\x^2=-9\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-9},\:x=-\sqrt{-9}$

$x=\sqrt{-9} = \sqrt{-1}\sqrt{9}=\sqrt{9}i=3i\\\\x=-\sqrt{-9}=-\sqrt{-1}\sqrt{9}=-\sqrt{9}i=-3i$

The solutions are:

$x=3i,\:x=-3i$

B. Two expressions are equivalent to each other if they represent the same value no matter which values we choose for the variables.

To factor $x^2+9$ :

First, multiply the constant in the polynomial by $i^2$ where $i^2$ is equal to -1.

$x^2+9i^2$

Since both terms are perfect squares, factor using the difference of squares formula

$a^2-i^2=(a+i)(a-i)$

$x^2+9=x^2+9i^2=\left(-3i+x\right)\left(3i+x\right)$

5 0

3 years ago

Find the length of the missing angle

amm1812

Answer: x=7.21

Step-by-step explanation: This is a right triangle- remember the Pythagorean theorem: a² + b² = c². To find leg a, you'd use a = √(c² - b²) **leg a would be the side that is labeled with 12. To find leg b, you'd use b = √(c² - a²) *this is the equation we need. To find the hypotenuse, or <em>c</em>, you'd use c = √(a² + b²). This equation works for right triangles only.

Hope this helps :)

4 0

3 years ago

6 3/5-2 4/5=?<br> Show how you got the answer

Viktor [21]

Hello there,

Okay so the question is asking us to subtract the two mixed numbers, 6 3/5 - 2 4/5, from each other...

The first step would be to change the mixed numbers to improper fractions:

6 3/5 as improper fraction would be 33/5
2 4/5 as improper fraction would be 14/5

Now we need to make sure that the denominators are the same and in this case they are.

Simply subtract 14 from 33 and you would get 19 and the denominator stays the same so you would get 19/5.

Now turn 19/5 into a mixed number and you would get 3 4/5

The answer to the 6 3/5 - 2 4/5 would be 3 4/5.

Hope I helped,

Amna

7 0

3 years ago

Please help me solve this

Tema [17]

Answer:

Step-by-step explanation:

I think it is she sped up and passed susana since it shows it in the graph.

5 0

3 years ago

V1 - 2 sin cos 0 = cos 0 - sin​

V1 - 2 sin cos 0 = cos 0 - sin