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stellarik [79]
3 years ago
6

Please help me out please

Mathematics
1 answer:
vivado [14]3 years ago
4 0
Use the scientific calculator from desmos hope that helps
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Complete the recursive formula of the arithmetic sequence 14, 30, 46, 62, ....
Sauron [17]

Answer:

They are adding 16 to the number each time. So the next number in the sequence would be 78.

Step-by-step explanation:

d(1) = 14

d(n) = d (n-1) + 16

14 + 16 = 30

30 + 16 = 46

46 + 16 = 62

62 + 16 = 78 and so on...

5 0
3 years ago
HELPPPPPPPPPPPPPPPPPPPPP
Tamiku [17]
$20/5 is $4, so for every increase in temperature, it's 4 (1°=$4). So 9*4=36. So your answer is B.
4 0
3 years ago
Read 2 more answers
Pls, help will give brainiest!!!!!!!!!!!
Zarrin [17]

Answer:

the first one

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Integrate e^x(sin(x) cos(x))
Karo-lina-s [1.5K]
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
Math question don't guess Please help
ICE Princess25 [194]
Option B is the answer to your question..!!
Hope this helps you!!!!

5 0
3 years ago
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