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3 years ago

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Kevin drew a diagonal inside a quadrilateral. He realizes the it always make 2 congruent isosceles triangleso. What kind of quad

rilateral did Kevin use to draw the triangles ?

1 answer:

olya-2409 [2.1K]3 years ago

7 0

Answer:

The square is the quadrilateral.

Step-by-step explanation:

A quadrilateral is a 2 dimensional geometric shape. It has four sides which may be or may not be equal.

If a quadrilateral is a square and we make its diagonal so that we get two triangles.

These two triangles are isosceles triangle as the two sides are equal and the square has four equal sides and the triangles are congruent to each other.

here, the triangles ABC and ADC are congruent and isosceles.

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Answer:

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Step-by-step explanation:

The given differential equation is

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Taking out common factors.

$\frac{dz}{dt}=(5z+2)e^{\sin t}\cos t$

Using variable separable method, we get

$\frac{dz}{5z+2}=e^{\sin t}\cos t dt$

Integrate both sides.

$\int\frac{dz}{5z+2}=\int e^{\sin t}\cos t dt$

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First solve LHS,

$I_1=\int\frac{dz}{5z+2}$

Substitute $5z+2=u$

$5\frac{dz}{du}=1$

$dz=\frac{1}{5}du$

$I_1=\frac{1}{5}\int\frac{du}{u}$

$I_1=\frac{1}{5}ln|u|+C_1$

$I_1=\frac{1}{5}ln|5z+2|+C_1$

Now, solve RHS,

$I_2=\int e^{\sin t}\cos t dt$

Substitute $\sin t=v$ ,

$\cos t\frac{dt}{dv}=1$

$\cos tdt=dv$

$I_2=\int e^{v}dv$

$I_2=e^{v}+C_2$

$I_2=e^{\sin t}+C_2$

Subtitle the values of I₁ and I₂ in equation (1).

$\frac{1}{5}ln|5z+2|+C_1=e^{\sin t}+C_2$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C_2-C_1$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C$

Therefore the general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$ .

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