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kodGreya [7K]
3 years ago
14

Lol im SimpForKirishima my account got baneed comeback kait

Mathematics
1 answer:
Softa [21]3 years ago
4 0

Answer:

Dats tuff

Step-by-step explanation:

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What is the answer to this inequality question 2(4x+1)<3(2x-3)
Nikitich [7]

Answer:

x < -7

Step-by-step explanation:

2(4x + 1) < 3(2x - 3)

Distributive Property:

8x + 8 < 6x - 6

Subtract on Both Sides:

8x + 8 < 6x - 6

- 6x       -6x

--------------------

2x + 8 < - 6

      -8    -8

---------------

2x < -14

Divide on Both Sides:

2x < -14

---     ---

2        2

x < -7

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AVprozaik [17]
Integers are whole numbers that range from negative infinity to positive infinity
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4 years ago
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Ilya [14]

Answer:

B. -11 + 7

Step-by-step explanation:

Minus a negative translates to add.

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4 0
3 years ago
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son4ous [18]

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9,801

Step-by-step explanation:

99*99=9,801

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Solve the equation on the interval [0,2π]
WARRIOR [948]
\bf 16sin^5(x)+2sin(x)=12sin^3(x)&#10;\\\\\\&#10;16sin^5(x)+2sin(x)-12sin^3(x)=0&#10;\\\\\\&#10;\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\&#10;-------------------------------\\\\&#10;2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x=&#10;\begin{cases}&#10;0\\&#10;\pi \\&#10;2\pi &#10;\end{cases}\\\\&#10;-------------------------------

\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0&#10;\\\\\\&#10;~~~~~~~~~~~~\textit{quadratic formula}&#10;\\\\&#10;\begin{array}{lcccl}&#10;& 8 sin^4& -6 sin^2(x)& +1\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array} &#10;\qquad \qquad &#10;sin(x)= \cfrac{ -  b \pm \sqrt {  b^2 -4 a c}}{2 a}&#10;\\\\\\&#10;sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}&#10;\\\\\\&#10;sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)=&#10;\begin{cases}&#10;\frac{1}{2}\\\\&#10;\frac{1}{4}&#10;\end{cases}

\bf \measuredangle x=&#10;\begin{cases}&#10;sin^{-1}\left( \frac{1}{2} \right)&#10;sin^{-1}\left( \frac{1}{4} \right)&#10;\end{cases}\implies \measuredangle x=&#10;\begin{cases}&#10;\frac{\pi }{6}~,~\frac{5\pi }{6}\\&#10;----------\\&#10;\approx~0.252680~radians\\&#10;\qquad or\\&#10;\approx~14.47751~de grees\\&#10;----------\\&#10;\pi -0.252680\\&#10;\approx 2.88891~radians\\&#10;\qquad or\\&#10;180-14.47751\\&#10;\approx 165.52249~de grees&#10;\end{cases}
3 0
3 years ago
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