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larisa [96]
3 years ago
6

What is the difference between 26 and 8

Mathematics
2 answers:
lbvjy [14]3 years ago
6 0

Answer:

the difference is 18

Step-by-step explanation:

Nady [450]3 years ago
4 0
18 because you subtract 8 from 26 and you get the difference
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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
Simplify the equation<br> 6 – 6w + 6w + 9 + 10w<br> Please I need help with this I'm stuck
Aneli [31]

Answer:

10w+15

Step-by-step explanation:

-6w+6w+10w=10w

6+9=15

hope this helps. Happy Holidays

7 0
3 years ago
What is the answer please help?
sammy [17]
I got d) not here. I solved it using my knowledge with angle relationships. I think the same rules apply with circles, but I am not 100% sure so please double check the math.

6 0
3 years ago
Find the area of the shaded region.
luda_lava [24]

Answer:

The area of the shaded region is 11.6\ cm^{2}

Step-by-step explanation:

we know that

The area of the shaded region is equal to the area of the sector of circle of angle 68.9 degrees minus the area of the isosceles triangle

step 1

Find the area of sector of the circle

The area of circle is equal to

A=\pi r^{2}

assume

\pi =3.14

r=9.28\ cm

substitute

A=(3.14)(9.28)^{2}

A=270.41\ cm^{2}

Remember that the area of a circle subtends a central angle of 360 degrees

so

using proportion Find out the area of a sector with a central angle of 68.90 degrees

Let

x -----> the area of a sector

270.41/360=x/68.90\\\\x=68.90*270.41/360\\\\x=51.75\ cm^{2}

step 2

Find the area of the isosceles triangle

Applying the law of sines

The area is equal to

A=(1/2)r^{2}sin(68.90)

we have

r=9.28\ cm

substitute

A=(1/2)(9.28)^{2}sin(68.90)=40.17\ cm^{2}

step 3

Find the area of the shaded region

51.75-40.17=11.58\ cm^{2}

Round to the nearest tenth

11.58=11.6\ cm^{2}

3 0
3 years ago
I NEED HELP ASAP BEFORE MARCH 26th <br> Find the Value of X
leva [86]

Answer:

45

Step-by-step explanation:

90+45=135 180-135 is 45

7 0
3 years ago
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