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RideAnS [48]
3 years ago
7

Find the exact values of the six trigonometric functions of θ if θ is in standard position and the terminal side of θ is in the

specified quadrant and satisfies the given condition.
III; parallel to the line
5y − 2x + 1 = 0
Mathematics
2 answers:
mixer [17]3 years ago
7 0
5y - 2x + 1 = 0.
5y = 2x-1.

y = 2/5 x - 2/5

this tells us that tan(theta) = 2/5
sin(theta)/sqrt(1-sin^2(theta)) = 2/5
x/sqrt(1-x^2)=2/5
x^2/(1-x^2)=4/25
25x^2=4-4x^2
29x^2=4
x^2=4/29
x = -2/sqrt(29) because quadrant III
cos(x) = sqrt(1-4/29) = -5/sqrt(29)


tan(theta)=2/5
cos(theta)=-5/sqrt(29)
sin(theta)=-2/sqrt(29)
cot(theta)=5/2
sec(theta)=-sqrt(29)/5
csc(theta)=-sqrt(29)/2
Pani-rosa [81]3 years ago
7 0
So the equation of the line is 5y -2x + 1 = 0.

Let's put that in standard form (by solving for y):

5y -2x + 1 = 0
5y = 2x - 1 ~ we added 2x and subtracted 1 from both sides
y =  \frac{2}{5}x - \frac{1}{5} ~ divide both sides by 5

So now we have it in the standard form which is y = mx + b.

The m gives us the slope. Recall that slope is rise over run.

So the slope is \frac{2}{5}, which means that you could draw a triangle with a rise of 2 and a run of 5.

That's almost enough information to find the trig values, but we still need to know the hypotenuse. Well, thanks to the magic of the Pythagorean Theorem:

hyp =  \sqrt{2^2 + 5^2} =  \sqrt{4+25} =  \sqrt{29}

Alright, now that we know all three sides, we can find the values.

The "rise" (vertical) or opposite: 2
The "run" (horozontial) or adjacent: 5
The hypotenuse: \sqrt{29}

Remember SOHCAHTOA. In order from that mnemonic: 

sin(θ) = opposite over hypotenuse = \frac{2}{\sqrt{29}}

cos(θ) = adjacent over hypotenuse = \frac{5}{\sqrt{29}}

tan(θ) = opposite over adjacent = \frac{2}{5}

Now the others (cosecant, secant, and cotangent) are just the reciprocals (upside-downs) of the others. Remember that cosecant comes first (since sine comes first in the "usuals".

csc(θ) = \frac{\sqrt{29}}{2}

sec(θ) = \frac{\sqrt{29}}{5}

cot(θ) = \frac{5}{2}

We don't have to 'solve' (approximate) that square root in any of the answers since the problem asked for the exact solution.
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Gnom [1K]

Answer:

The answer would be A.

Hope this helped :D

7 0
2 years ago
A seamstress is paid $9.55 for every pair of pants made. how many pants would have to be made to receive $525.00 a week?
Svetlanka [38]
Let the number of pants have to be made be x.

So,

9.55 * x = 525.00

x = \frac{525}{9.55} ≈ 54.97 

Thus, the number of pants have to be made are 54.
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3 years ago
8. Determine the x- and y-intercepts of -12a - 8y = 24.<br> i need help
Doss [256]

Step-by-step explanation: First you convert the equation into standard form, which is y=mx+b . Then graph it. When it is in the form y=mx+b it is easier to graph. Then you can see the intercepts on the graph and come up with your answer.

Hope this helps!

3 0
2 years ago
The population of Oak Forest is increasing at a rate of 4% per year. If the population is 74,145 today, what will it be in three
Shtirlitz [24]

Answer:

83,403

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Hope this helps!

4 0
3 years ago
Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]

The left and right endpoints of the i-th subinterval, respectively, are

\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8

r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8

for 1\le i\le4, and the respective midpoints are

m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8

  • Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}

  • Midpoint rule

We approximate the area for each subinterval by

M_i=f(m_i)(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}

  • Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial p_i(x), where

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It so happens that the integral of p_i(x) reduces nicely to the form you're probably more familiar with,

S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))

Then the integral is approximately

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0
3 years ago
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