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3 years ago

Hey, can somebody please help me with this? I think it might be A but I’m not completely sure. Can someone give their thoughts?

Mathematics

1 answer:

Ratling [72]3 years ago

5 0

Answer:

I Believe The Answer Is B Plz Tell Me If I Am Right

Step-by-step explanation:

The Chart Shows More Predictability On Chart B Since There Is Less Outliers

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Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B

telo118 [61]

In order to prove

$\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)$

Let's write both sides in terms of $\sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x)$ only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

$\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)$

and

$\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)$

So, their multiplication is

$\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)$

So, the left hand side simplifies to

$\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now, on with the right hand side. We have

$1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now simply make this expression one fraction:

$1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

And as you can see, the two sides are equal.

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4 years ago

Number of different types of atoms in the universe-

SashulF [63]

Their are many different types of atoms in the universe. It can include the proton, neutron, and the electron. Also, the quark is a particle but is not found alot.

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3 years ago

Which of the following expressions would give you the greatest quotient (answer to the division problem)? PLEASEE ITS THE LAST Q

Alja [10]

Help me or my family and my friends are in my

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3 years ago

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If sin theta = 2/3 and sex theta < 0 , find cos theta and tan theta

FromTheMoon [43]

Answer:

$\displaystyle cos\theta=-\frac{\sqrt{5}}{3}$

$\displaystyle tan\theta=-\frac{2\sqrt{5}}{5}$

Step-by-step explanation:

<u>Trigonometric Formulas</u>

To solve this problem, we must recall some basic relations and concepts.

The main trigonometric identity relates the sine to the cosine:

$sin^2\theta+cos^2\theta=1$

The tangent can be found by

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

The cosine and the secant are related by

$\displaystyle cos\theta=\frac{1}{sec\theta}$

They both have the same sign.

The sine is positive in the first and second quadrants, the cosine is positive in the first and fourth quadrants.

The sine is negative in the third and fourth quadrants, the cosine is negative in the second and third quadrants.

We are given

$\displaystyle sin\theta=\frac{2}{3}$

Find the cosine by solving

$sin^2\theta+cos^2\theta=1$

$\displaystyle \left(\frac{2}{3}\right)^2+cos^2\theta=1$

$\displaystyle cos^2\theta=1-\left(\frac{2}{3}\right)^2=1-\frac{4}{9}=\frac{5}{9}$

$\displaystyle cos\theta=\sqrt{\frac{5}{9}}=-\frac{\sqrt{5}}{3}$

$\boxed{\displaystyle cos\theta=-\frac{\sqrt{5}}{3}}$

We have placed the negative sign because we know the secant ('sex') is negative and they both have the same sign.

Now compute the tangent

$\displaystyle tan\theta=\frac{sin\theta}{cos\theta}=\frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}}=-\frac{2}{\sqrt{5}}$

Rationalizing

$\displaystyle tan\theta=-\frac{2}{\sqrt{5}}=-\frac{2\sqrt{5}}{5}$

$\boxed{\displaystyle tan\theta=-\frac{2\sqrt{5}}{5}}$

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Tatiana [17]

Point symmetry and line symmetry are kinds of symmetry

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