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2 years ago

Help for good luck and a lot of money

Mathematics

2 answers:

AVprozaik [17]2 years ago

7 0

How many years of good luck are we talking about?

It would be 12.2 lbs per package

Since 24.4/2 = 12.2

It would be 97.6 lbs. for 8 packages

122 lbs. would equal out to be 10 packages.

Alex777 [14]2 years ago

5 0

1= 12.2
8=97.6
and for the last one the answer is 10 :))

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Use the relationship between the angles in the figure to answer the question. Which equation can be used to find the value of x?

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The sum of all the angles made by a straight line is 180°. Then the value of x is 61°.

<h3>What is an angle?</h3>

Angle is the space between the line or the surface that meets. And the angle is measured in degree. For complete 1 rotation, the angle is 360 degrees.

The angles are shown on the diagram.

The angles are 52° and 67°. And let the other angle be x.

We know that the sum of all the angles made by a straight line is 180°.

Then we have

x + 52° + 67° = 180°

On simplifying, we have the value of x.

x = 61°

Thus, the value of x is 61°.

More about the angles link is given below.

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3 years ago

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Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

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