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Strike441 [17]
3 years ago
12

Figure ABCD is a rhombus where the m∠ABC = 84 and m∠ABE = 3x − 6. Solve for x. Rhombus ABCD with diagonals AC and BD and point E

as the point of intersection of the diagonals. 16 30 32 62

Mathematics
2 answers:
sladkih [1.3K]3 years ago
5 0

Answer:

16

Step-by-step explanation:

bazaltina [42]3 years ago
4 0

Answer:

<u> x = 16</u>

Step-by-step explanation:

See the attched figure

We should know that, one of the properties of the rhombus is the diagonals bisect the angles of the rhombus.

Given:

m∠ABC = 84  and m∠ABE = 3x − 6

So, m∠ABE = 0.5 * m∠ABC = 0.5 * 84 = 42

∴ 3x - 6 = 42

Solve for x

3x = 42 + 6 = 48

x = 48/3 = 16

<u>∴ x = 16</u>

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Supposing that |y|, we have \tan^{-1}u=y, from which it follows that

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So we can write the ODE as

\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3

which is linear in u. Multiplying both sides by e^{x^2}, we have

e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}
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Integrate both sides with respect to x:

\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx
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Substitute t=x^2, so that \mathrm dt=2x\,\mathrm dx. Then

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f=t\implies\mathrm df=\mathrm dt
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y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)
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