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3 years ago

Put the brackets in to make this equation acurate 9+7x3÷10-1=2

Mathematics

2 answers:

liberstina [14]3 years ago

6 0

Question is wrong.......any bracket can not solve this problem

Amanda [17]3 years ago

4 0

The question has to be wrong i tried putting brackets in for each and i never came out with 2

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Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the<em> </em><u><em>Weierstrass substitution</em></u>, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$

Hence, we showed $\sin(x) \text { and } \cos(x)$

======================================================

$5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]$

Solving

$5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3$

$\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies \dfrac{5-5t^2 -24t}{1+t^2}= 3$

$\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0$

$t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} = \dfrac{3\pm \sqrt{10}}{-2}$

$t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}$

Just note that

$\tan\left(\dfrac{x}{2}\right) = \dfrac{3\pm 8\sqrt{10}}{-2}$

and $\tan\left(\dfrac{x}{2}\right)$ is not defined for $x=k\pi , k\in\mathbb{Z}$

6 0

3 years ago

How do you get over a boy who has left you 6 times..?

PolarNik [594]

Who has left me for 6 times man

seriously

I wouldn't get into that boy firstly

<em>Lol</em>

4 0

3 years ago

A gardener has 520 feet of fencing to fence in a rectangular garden. One side of the garden is bordered by a river and so it doe

Bond [772]

The shortest side is 130 feet, the longest side is 260 feet and the greatest possible area is 33800 square feet

<h3>What dimensions would guarantee that the garden has the greatest possible area?</h3>

The given parameter is

Perimeter, P = 520 feet

Represent the shorter side with x and the longer side with y

One side of the garden is bordered by a river:

So the perimeter is:

P = 2x + y

Substitute P = 520

2x + y = 520

Make y the subject

y = 520 - 2x

The area is

A = xy

Substitute y = 520 - 2x in A = xy

A = x(520 - 2x)

Expand

A = 520x - 2x^2

Differentiate

A' = 520 - 4x

Set to 0

520 - 4x = 0

Rewrite as:

4x= 520

Divide by 4

x= 130

Substitute x= 130 in y = 520 - 2x

y = 520 - 2 *130

Evaluate

y = 260

The area is then calculated as:

A = xy

This gives

A = 130 * 260

Evaluate

A = 33800

Hence, the shortest side is 130 feet, the longest side is 260 feet and the greatest possible area is 33800 square feet