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Anuta_ua [19.1K]
3 years ago
10

X, add 6, then times by 3

Mathematics
1 answer:
Musya8 [376]3 years ago
4 0

Answer:

(x+6)*3

Step-by-step explanation:

You will need to do this equation simply; and in order for that matter.

So you need to use the brackets so you know what comes first.

Remember BIDMAS or BODMAS

BIDMAS:                  BODMAS:

Brackets                  Brackets

Indices                     Order       ---   (These mean the same thing)

Divide                      Divide

Multiply                    Multiply

Addition                   Addition

Subtraction              Subtraction

As you can see the brackets come first, so you do what is INSIDE the brackets before anything. (x+6)

To finish, you will need to multiply, but as it is an unknown, you don't know what to multiply the 3 by.

<u>So x, add 6, then times by 3 = (x+6) *3</u>

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Using a trigonometric identity, it is found that the values of the cosine and the tangent of the angle are given by:

  • \cos{\theta} = \pm \frac{2\sqrt{2}}{3}
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<h3>What is the trigonometric identity using in this problem?</h3>

The identity that relates the sine squared and the cosine squared of the angle, as follows:

\sin^{2}{\theta} + \cos^{2}{\theta} = 1

In this problem, we have that the sine is given by:

\sin{\theta} = \frac{1}{3}

Hence, applying the identity, the cosine is given as follows:

\cos^2{\theta} = 1 - \sin^2{\theta}

\cos^2{\theta} = 1 - \left(\frac{1}{3}\right)^2

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\cos^2{\theta} = \frac{8}{9}

\cos{\theta} = \pm \sqrt{\frac{8}{9}}

\cos{\theta} = \pm \frac{2\sqrt{2}}{3}

The tangent is given by the sine divided by the cosine, hence:

\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}

\tan{\theta} = \frac{\frac{1}{3}}{\pm \frac{2\sqrt{2}}{3}}

\tan{\theta} = \pm \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}

\tan{\theta} = \pm \frac{\sqrt{2}}{4}

More can be learned about trigonometric identities at brainly.com/question/24496175

#SPJ1

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Cho f(t)=(t-4)u(t-2)phep bien doi laplace la<br> F(s)=e^-2s/s^2-2e^-2s/s62
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Step-by-step explanation:

cho f(t)=(t-4)u(t-2)phep bien doi laplace la

F(s)=e^-2s/s^2-2e^-2s/s62

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Answer:

See below  

Step-by-step explanation:

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The field lines spread out radially from the centre of the point. They are represented by arrows pointing in the direction that a positive charge would move if it were in the field.

Opposite charges attract, so the field lines point toward the centre of the particle.

For an isolated negative particle, the field lines would look like those in Figure 1 below.

If two negative charges are near each other, as in Figure 2, the field lines still point to the centre of charge.

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We can make a similar statement about appositive charge approaching from the left.

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From what i see it could be
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