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german
3 years ago
8

Ex 3.6 6. find the area enclosed between the curve y= -2x²-5x+3 and the x-axis

Mathematics
2 answers:
Xelga [282]3 years ago
6 0
When y=0,

-2{ x }^{ 2 }-5x+3=0\\ \\ 2{ x }^{ 2 }+5x-3=0\\ \\ \left( 2x-1 \right) \left( x+3 \right) =0

\\ \\ \therefore \quad x=\frac { 1 }{ 2 } \\ \\ \therefore \quad x=-3

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

\int _{ -3 }^{ \frac { 1 }{ 2 }  }{ -2{ x }^{ 2 } } -5x+3dx

\\ \\ ={ \left[ -\frac { { 2x }^{ 2+1 } }{ 2+1 } -\frac { 5{ x }^{ 1+1 } }{ 1+1 } +3x \right]  }_{ -3 }^{ \frac { 1 }{ 2 }  }

\\ \\ ={ \left[ -\frac { 2{ x }^{ 3 } }{ 3 } -\frac { 5{ x }^{ 2 } }{ 2 } +3x \right]  }_{ -3 }^{ \frac { 1 }{ 2 }  }

\\ \\ \\ =\left\{ -\frac { 2 }{ 3 } { \left( \frac { 1 }{ 2 }  \right)  }^{ 3 }-\frac { 5 }{ 2 } { \left( \frac { 1 }{ 2 }  \right)  }^{ 2 }+3\left( \frac { 1 }{ 2 }  \right)  \right\} -\left\{ -\frac { 2 }{ 3 } { \left( -3 \right)  }^{ 3 }-\frac { 5 }{ 2 } { \left( -3 \right)  }^{ 2 }+3\left( -3 \right)  \right\}

\\ \\ \\ =-\frac { 2 }{ 3 } \cdot \frac { 1 }{ 8 } -\frac { 5 }{ 2 } \cdot \frac { 1 }{ 4 } +\frac { 3 }{ 2 } -\left\{ -\frac { 2 }{ 3 } \left( -27 \right) -\frac { 5 }{ 2 } \cdot 9-9 \right\}

\\ \\ =-\frac { 2 }{ 24 } -\frac { 5 }{ 8 } +\frac { 3 }{ 2 } -\left\{ \frac { 54 }{ 3 } -\frac { 45 }{ 2 } -9 \right\}

\\ \\ =-\frac { 2 }{ 24 } -\frac { 15 }{ 24 } +\frac { 36 }{ 24 } -\frac { 54 }{ 3 } +\frac { 45 }{ 2 } +9

\\ \\ =\frac { 19 }{ 24 } -\frac { 54 }{ 3 } +\frac { 45 }{ 2 } +\frac { 18 }{ 2 } \\ \\ =\frac { 19 }{ 24 } -\frac { 54 }{ 3 } +\frac { 63 }{ 2 }

\\ \\ =\frac { 343 }{ 24 }

Answer: 343/24 units squared.
butalik [34]3 years ago
6 0
Y = -2x² - 5x + 3
-2x² - 5x + 3 = 0
x = <u>-(-5) +/- √(-5² - 4(-2)(3))</u>
                     2(-2)
x = <u>5 +/- √(25 + 24)</u>
                -4
x = <u>5 +/- √39
</u>           -4<u>
</u>x = <u>5 +/- 6.244997998398398
</u>                      -4<u>
</u>x = <u>5 + 6.244997998398398 </u>  x = <u>5 - 6.244997998398398</u>                     
                       -4                                             -4<u>
</u>x = <u>11.624997998398398</u>       x = <u>-1.244997998398398
</u>                    -4                                             -4<u>
</u>x = -2.8112494995996           x = 0.3112494995996<u>
</u>-----------------------------------------------------------------------------------------------
y = -2x² - 5x + 3                      y = -2x² - 5x + 3
y = -2(-3)² - 5(-3) + 3               y = -2(0.3)² - 5(0.3) + 3
y = -2(9) + 15 + 3                    y = -2(0.09) - 0.15 + 3
y = -18 + 15 + 3                      y = -0.18 - 0.15 + 3
y = -3 + 3                                y = -0.33 + 3
y = 0                                       y = 2.67
-----------------------------------------------------------------------------------------------
(x, y) = (-3, 0)                          (x, y) = (0.3, 2.67)
<u />
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Find x and the measure of each side.
Anna71 [15]

Answer:

x=27.5

FG=72.5

GH=60

HF=47.5

Step-by-step explanation:

Sum of angles in a triangle=180

Therefore, FG+GH+HF=180

Substituting the values we have,

(3x - 10) + (2x+5) +(x+20) = 180

Collecting like terms

3x+2x+x = 180 +10 - 5 - 20

6x = 165

X= 27.5

FG = (3x - 10) = 3(27.5) - 10

FG= 72.5

GH= (2x+5) = 2(27.5)+5

GH=60

HF= x+20 =27.5+20

HF=47.5

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1 year ago
A wheel spins at 300 revolutions per minute. What is the angular velocity of the wheel, in radians per second? Round the answer
Ostrovityanka [42]

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\cfrac{300~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{2\pi ~rad}{~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{60secs}\implies \cfrac{(300)(2\pi )rad}{60secs}\implies 10\pi ~\frac{rad}{secs}\approx 31.42~\frac{rad}{secs}

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Natasha2012 [34]
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3 years ago
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