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tangare [24]
3 years ago
7

Help me with this please

Mathematics
1 answer:
olga2289 [7]3 years ago
4 0

Answer:

The one you chose

Step-by-step explanation:

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What's the answer??? I
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If you want to know the answer you need to give us the question.
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4 years ago
C=0.04x^2-8.504x+25302 how do i do this?
erma4kov [3.2K]
Solving for x.

Since x<span> is on the right-hand side of the </span>equation<span>, switch the sides so it is on the left-hand side of the </span>equation<span>.
</span><span>0.04<span>x2</span>−8.504x+25302=c

</span>Simplify the quadratic and set<span> the right side equal to </span><span>0.
</span><span>0.04<span>x2</span>−8.504x+25302−c=0

</span>Use the standard form of the quadratic <span>(<span><span>a<span>x2</span>+bx+c</span>)</span></span><span> to find </span><span>a,</span><span>b,</span><span> and </span>c<span> for this quadratic.
</span><span>a=0.04,</span><span>b=−8.504,</span><span>c=25302−1c

</span>Use the quadratic formula<span> to find the </span>solutions<span>.
</span><span>x=<span>−b ± </span></span>√<span><span> <span><span>b2</span>−4ac /</span> </span><span>2a

</span></span>Substitute in the values of <span><span>a=0.04</span>,</span><span><span>b=−8.504</span>,</span><span> and </span><span><span>c=25302−1c</span>.
</span>x=<span>−(−8.504) ±</span>√<span><span>(−8.504<span>)2</span>−4(0.04)(25302−1c)/</span><span>2(0.04)
</span></span>
<span>Simplify.
</span>x=<span>8.504 ±</span>√<span><span>(−8.504<span>)2</span>−4(0.04)(25302−1c)/</span><span>2(0.04)
</span></span>
Simplify the section inside the radical<span>.
</span>x=<span>8.504 ± </span>√<span><span><span>−3976.001984+0.16c</span>/</span><span>2(0.04)
</span></span>
Simplify the denominator<span> of the </span>quadratic formula<span>.
</span>x=<span>8.504 ±</span>√<span><span>−3976.001984+0.16c/</span>0.08
</span>

Answer:
x=8.504 ±√−3976.001984+0.16c/0.08


6 0
3 years ago
Integration questions .
dlinn [17]
<h2>1)</h2>

\\\\\ \textbf{a)}\\\\~~~\displaystyle \int (6x- \sin 3x) ~ dx\\\\=6\displaystyle \int x ~ dx - \displaystyle \int \sin 3x ~ dx\\\\=6 \cdot \dfrac{x^2}2 - \dfrac 13 (- \cos 3x) +C~~~~~~~~~~~;\left[\displaystyle \int x^n~ dx = \dfrac{x^{n+1}}{n+1}+C,~~~n \neq -1\right]\\\\ =3x^2 +\dfrac{\cos 3x}3 +C~~~~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \sin (mx) ~dx = -\dfrac 1m ~ (\cos mx)+C \right]\\

\textbf{b)}\\\\~~~~\displaystyle \int(3e^{-2x} +\cos (0.5 x)) dx\\\\=3\displaystyle \int e^{-2x} ~dx+ \displaystyle \int \cos(0.5 x) ~dx\\\\\\=-\dfrac 32 e^{-2x} + \dfrac 1{0.5} \sin (0.5 x) +C~~~~~~~~~~~~~~;\left[\displaystyle \int e^{mx}~dx = \dfrac 1m e^{mx} +C \right]\\\\\\=-\dfrac 32 e^{-2x} + 2 \sin(0.5 x) +C~~~~~~~~~~~~~~~~~;\left[\displaystyle \int \cos(mx)~ dx  = \dfrac 1m \sin(mx) +C\right]\\\\\\=-1.5e^{-2x} +2\sin(0.5x) +C

<h2>2)</h2>

\textbf{a)}\\\\y = \displaystyle \int \cos(x+5) ~ dx\\\\\text{Let,}\\\\~~~~~~~u = x+5\\\\\implies \dfrac{du}{dx} = 1+0~~~~~~;[\text{Differentiate both sides.}]\\\\\implies \dfrac{du}{dx} = 1\\\\\implies du = dx\\\\\text{Now,}\\\\y= \displaystyle \int \cos u ~ du\\\\~~~= \sin u +C\\\\~~~=\sin(x+5) + C

\textbf{b)}\\\\y = \displaystyle \int 2(5x-3)^4 dx\\\\\text{Let,}\\~~~~~~~~u = 5x-3\\\\\implies \dfrac{du}{dx} = 5~~~~~~~~~~;[\text{Differentiate both sides}]\\\\\implies dx = \dfrac{du}5\\\\\text{Now,}\\\\y = 2\cdot \dfrac 1  5 \displaystyle \int u^4 ~ du\\\\\\~~=\dfrac 25 \cdot \dfrac{u^{4+1}}{4+1} +C\\\\\\~~=\dfrac 25 \cdot \dfrac{u^5}5+C\\\\\\~~=\dfrac{2u^5}{25}+C\\\\\\~~=\dfrac{2(5x-3)^5}{25}+C

<h2>3)</h2>

\textbf{a)}\\\\y =  \displaystyle \int xe^{3x} dx\\\\\text{We know that,}\\\\ \displaystyle \int  (uv) ~dx = u  \displaystyle \int  v ~ dx -  \displaystyle \int \left[ \dfrac{du}{dx} \displaystyle \int ~ v ~ dx \right]~ dx\\\\\text{Let}, u =x~ \text{and}~ v=e^{3x}  .\\\\y=  \displaystyle \int xe^{3x} ~dx\\\\\\~~=  x\displaystyle \int e^{3x} ~ dx -  \displaystyle \int  \left[\dfrac{d}{dx}(x)  \displaystyle \int  e^{3x}~ dx \right]~ dx\\\\\\

  =x\displaystyle \int e^{3x}~ dx  - \displaystyle \dfrac 13 \int \left(e^{3x} \right)~ dx\\\\\\=\dfrac{xe^{3x}}3 - \dfrac 13 \cdot \dfrac{ e^{3x}}3+C\\\\\\= \dfrac{xe^{3x}}{3}- \dfrac{e^{3x}}{9}+C\\\\\\=\dfrac{3xe^{3x}}{9}- \dfrac{e^{3x}}9 + C\\\\\\= \dfrac 19e^{3x}(3x-1)+C

 

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