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3 years ago

How would you set up this problem to find the height of the sail using reciprocal functions?

Mathematics

1 answer:

AVprozaik [17]3 years ago

5 0

Remark

I'm not sure I know what you mean, but I will answer trying to interpret it so it fits the description of the question.

Step One: Set up the equation. The normal way of solving this would be to use the Cos function.

Cos(theta) = Adjacent / hypotenuse = x/8 Multiply both sides by 8

8*cos(theta) = x Theta = 35

8*cos(35) = x This is about as simple a way as is possible.

6.553 = the height of the sale.

Step Three: try using the reciprocal

1 / cos(35)= hypotenuse / adjacent

1/0.8191 = 8/x cross multiply

x = 0.8191 * 8

x = 6.553

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4 years ago

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Can someone plz help me with this!

asambeis [7]

Answer:

(arranged from top to bottom)

System #3, where x=6

System #1, where x=4

System #7, where x=3

System #5, where x=2

System #2, where x=1

Step-by-step explanation:

System #1: x=4

$2x+y=10\\x-3y=-2$

To solve, start by isolating your first equation for y.

$2x+y=10\\y=-2x+10$

Now, plug this value of y into your second equation.

$x-3(-2x+10)=-2\\x+6x-30=-2\\7x=28\\x=4$

System #2: x=1

$x+2y=5\\2x+y=4$

Isolate your second equation for y.

$2x+y=4\\y=-2x+4$

Plug this value of y into your first equation.

$x+2(-2x+4)=5\\x+(-4x)+8=5\\x-4x+8=5\\-3x=-3\\x=1$

System #3: x=6

$5x+y=33\\x=18-4y$

Isolate your first equation for y.

$5x+y=33\\y=-5x+33$

Plug this value of y into your second equation.

$x=18-4(-5x+33)\\x=18+20x-132\\-19x=-114\\x=6$

System #4: all real numbers (not included in your diagram)

$y=13-2x\\8x+4y=52$

Plug your value of y into your second equation.

$8x+4(13-2x)=52\\8x+52-8x=52\\0=0$

all real numbers are solutions

System #5: x=2

$x+3y=5\\6x-y=11$

Isolate your second equation for y.

$6x-y=11\\-y=-6x+11\\y=6x-11$

Plug in your value of y to your first equation.

$x+3(6x-11)=5\\x+18x-33=5\\19x=38\\x=2$

System #6: no solution (not included in your diagram)

$2x+y=10\\-6x-3y=-2$

Isolate your first equation for y.

$2x+y=10\\y=-2x+10$

Plug your value of y into your second equation.

$-6x-3(-2x+10)=-2\\-6x+6x-30=-2\\-30=-2$

no solution

System #7: x=3

$y=10+x\\2x+3y=45$

Plug your value of y into your second equation.

$2x+3(10+x)=45\\2x+30+3x=45\\5x=15\\x=3$

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3 years ago

The question is...................

Norma-Jean [14]

B is dilated by a factor of 5

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3 years ago

<img src="https://tex.z-dn.net/?f=9%5C%3A%20%20%5C%3A%20%20%5Cfrac%7B%20%5Csin%28%20%5Ctheta%29%20%7D%7B1%20%2B%20%20%5Ccos%28%2

PIT_PIT [208]

Option (b) is your correct answer.

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given Trigonometric expression is

$\rm :\longmapsto\:\dfrac{sin\theta }{1 + cos\theta }$

So, on rationalizing the denominator, we get

$\rm \: = \: \dfrac{sin\theta }{1 + cos\theta } \times \dfrac{1 - cos\theta }{1 - cos\theta }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}}$

So, using this, we get

$\rm \: = \: \dfrac{sin\theta (1 - cos\theta )}{1 - {cos}^{2}\theta }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this identity, we get

$\rm \: = \: \dfrac{sin\theta (1 - cos\theta )}{{sin}^{2}\theta }$

$\rm \: = \: \dfrac{1 - cos\theta }{sin\theta }$

Hence, 

$\\ \red{\rm\implies \:\boxed{\tt{ \rm \:\dfrac{sin\theta }{1 + cos\theta } = \: \dfrac{1 - cos\theta }{sin\theta } }}} \\$

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3 years ago

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