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Tju [1.3M]
3 years ago
14

Which of the following is not a perfect square? OA, 1 OB) 2 OC) 4 OD) 9

Mathematics
1 answer:
Burka [1]3 years ago
5 0
B) 2
This is so because no number to the power of 2 has the result of 2

The other do
2x2 = 4
1x1 = 1
3x3 = 9
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x(x² - 12x + 35)

x(x-5)(x-7)
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Solve x+4=10 A.x=2 B.x=3 C.x=6 D.x=14
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C. x = 6

Step-by-step explanation:

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1. If you invest $500 in an account that earns 3.5% annually compounded monthly. How
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Integrate e^x(sin(x) cos(x))
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I=\int e^x(\sin(x)\cos(x))dx=\int e^x(\frac{1}{2}\sin(2x))dx=\frac{1}{2}\int e^x\sin(2x)dx

\text{If }u=\sin(2x)\to du=2\cos(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
\text{Using }\int u\,dv=uv-\int v\,du:\\\\
I=\frac{1}{2}(e^x\sin(2x)-\int e^x(2\cos(2x))dx)\\\\
2I=e^x\sin(2x)-2\underbrace{\int e^x\cos(2x)dx}_{I_2}

Looking for I_2:

\text{If}~u=\cos(2x)\to du=-2\sin(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
I_2=e^x\cos(2x)-\int e^x(-2\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(\sin(2x))dx\\\\  I_2=e^x\cos(2x)+2\int e^x(2\sin(x)\cos(x))dx\\\\ I_2=e^x\cos(2x)+4\int e^x(\sin(x)\cos(x))dx=e^x\cos(2x)+4I

Replacing:

2I=e^x\sin(2x)-2I_2\iff\\\\2I=e^x\sin(2x)-2(e^x\cos(2x)+4I)\iff\\\\
2I=e^x\sin(2x)-2e^x\cos(2x)-8I\iff\\\\
10I=e^x\sin(2x)-2e^x\cos(2x)\iff\\\\
I=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))\\\\
\boxed{\int e^x(\sin(x)\cos(x))dx=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))+C}
6 0
3 years ago
Please help . I’ll mark you as brainliest if correct!
geniusboy [140]

u4i4irjrrkrkeoeneoenr

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3 years ago
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