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3 years ago

What proves m ∆ADB and ∆ADC are congruent.

Mathematics

1 answer:

AURORKA [14]3 years ago

6 0

Answer: They're the same angles/measurements.

Step-by-step explanation: Look closely at both the angles. If you measure them, they will be the same because both angles look the same, but ΔADC's side is flipped.

Let me know if you have any questions.

~ Lily, from Brainly.

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Please help me please !!

sweet [91]

Answer:

Acute

Step-by-step explanation:

7² + 9² ? 11²

49 + 81 ? 121

130 is greater than 121, so it is an acute triangle

8 0

3 years ago

Find the value of x if 8∧3x/2∧10 = 4∧2x/16

solniwko [45]

Answer:

Solve for

by cross multiplying.

Exact Form:

6/5

Decimal Form:

1.2

Mixed Number Form:

1 1/5

6 0

3 years ago

I need help.Thanks!

IgorLugansk [536]

Answer:

35×14-16

Step-by-step explanation:

times means × and minus means -

3 0

3 years ago

Write ab in terms of a and b

baherus [9]

Answer:

a multiple b

Step-by-step explanation:

a x b = ab

eg. a = 5 , b = 2

ab = 5x2

8 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