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Cloud [144]
3 years ago
14

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal plac

es where appropriate. If there is no solution, enter NO SOLUTION.) 5 sin(2θ) − 2 sin(θ) = 0

Mathematics
1 answer:
coldgirl [10]3 years ago
5 0

Answer:

  x = {kπ, arccos(1/5) +2kπ, 2kπ -arccos(1/5)}

Step-by-step explanation:

The double-angle trig identity for sine is useful:

  5(2sin(θ)cos(θ)) -2sin(θ) = 0

  2sin(θ)(5cos(θ) -1) = 0

This has solutions that make the factors zero:

  θ = arcsin(0) = kπ

and ...

  cos(θ) = 1/5

  θ = arccos(1/5) +2kπ . . . . or . . . . 2kπ -arccos(1/5)

_____

Some numerical values are shown on the graph attached. values for multiples of pi are ...

  {..., -12.566, -9.425, -6.283, -3.142, 0, 3.142, 6.283, 9.425, 12.566, ...}

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Softa [21]

Answer:

using quaderatic expression solving problems of algebraic expressions.

Step-by-step explanation:

no1.

take the smallest side to be x

and the longest side to be x-1/3

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6 0
2 years ago
The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p
Juliette [100K]

Check the picture below.


based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,


\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)


\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}


now, we can plug those values on A = (1/2)bh,


\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5

7 0
3 years ago
Meg initially has 3 hours of pop music and 2 hours of classical music in her collection. Every month onwards, the hours of pop m
zlopas [31]

Answer:  Option 'd' is correct.

Step-by-step explanation:

Since we have given that

Number of hours of pop music = 3

Number of hours of classical music = 2

According to question, Every month onwards, the hours of pop music in her collection is 5% more than what she had the previous month. Her classical music does not change.

Rate of increment = 5% = 0.05

Let the number of months be 'x'.

So, our required function becomes,

f(x)=3(1+0.05)x+2\\\\f(x)=3(1.05)x+2

Hence, Option 'd' is correct.

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Realy need help if you help thank you so much
stepladder [879]

Answer:

B

Step-by-step explanation:

Rate, in this case, is mph which can be written as miles/hours.

HTH :)

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The two-dimensional figures would be;

Rectangles and Triangles

Rectangles would be front, back and bottom

Triangles would be the 2 right and left sides. 
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