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kicyunya [14]
3 years ago
6

Solve 6 sin((π/5)x)=5 for the four smallest positive solutions

Mathematics
1 answer:
cupoosta [38]3 years ago
3 0

Answer:

1.568, 3.432, 11.568, 13.432

Step-by-step explanation:

Divide equation 6\sin\left(\dfrac{\pi }{5}x\right)=5 by 6:

\sin\left(\dfrac{\pi }{5}x\right)=\dfrac{5}{6}.

Then

\dfrac{\pi }{5}x=(-1)^k\arcsin\left(\dfrac{5}{6}\right)+\pi k,\ k\in Z,

x=(-1)^k\dfrac{5}{\pi }\arcsin\left(\dfrac{5}{6}\right)+5k,\ k\in Z.

Since \arcsin\left(\dfrac{5}{6}\right)\approx 56^{\circ}\approx \dfrac{\pi }{3.2},

four smallest positive solutions are

1. \dfrac{5}{3.2}\approx 1.568,\ k=0;

2. \dfrac{-5}{3.2}+5\approx 3.432,\ k=1;

3. \dfrac{5}{3.2}+10\approx 11.568,\ k=2;

4. \dfrac{-5}{3.2}+15\approx 13.432,\ k=3.

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If GF is a midsegment of CDE, find CD.<br><br> A. 3.4<br><br>B. 6.8 <br><br>C. 13.6 <br><br>D. 14
bogdanovich [222]

Answer:

C


Step-by-step explanation:

Triangle CGF and triangle CED are similar. Hence, the ratio of their corresponding sides are equal. Thus we can write:

\frac{5x+4}{2x+3}=\frac{CD}{3x+0.8}

<em>We can now cross multiply and solve for CD:</em>

\frac{5x+4}{2x+3}=\frac{CD}{3x+0.8}\\(5x+4)(3x+0.8)=(2x+3)(CD)\\15x^2+4x+12x+3.2=(2x+3)(CD)\\15x^2+16x+3.2=(2x+3)(CD)\\CD=\frac{15x^2+16x+3.2}{2x+3}

Since GF is a midsegment of CDE, CD is double of CF. So we can write:

CD=2CF\\\frac{15x^2+16x+3.2}{2x+3}=2(3x+0.8)\\\frac{15x^2+16x+3.2}{2x+3}=6x+1.6\\15x^2+16x+3.2=(6x+1.6)(2x+3)\\15x^2+16x+3.2=12x^2+18x+3.2x+4.8\\15x^2+16x+3.2=12x^2+21.2x+4.8\\3x^2-5.2x-1.6=0

<em>By using quadratic formula \frac{-b+-\sqrt{b^2-4ac} }{2a} and with a=3, b= -5.2, and c= -1.6, we find the value of x to be:</em>

\frac{5.2+-\sqrt{(-5.2)^2-4(3)(-1.6)} }{2(3)}=2


<em>Since the expression for CD is \frac{15x^2+16x+3.2}{2x+3} , we plug in x=2 into this expression to find value of CD:</em>

\frac{15(2)^2+16(2)+3.2}{2(2)+3}=13.6

The correct answer is C

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Answer:

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Step-by-step explanation:

5x + 8y = 139

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7 0
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