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4 years ago

Triangle ABC has vertices at A(-5, 4), B(4, 1), and C(1, -8). Choose the terms below which correctly describe this triangle:

Mathematics

1 answer:

Anton [14]4 years ago

3 0

Answer:

An ISOSCELES TRIANGLE

Step-by-step explanation:

Given a triangle ABC with vertices at A(-5, 4), B(4, 1), and C(1, -8), to know the type of triangle this is, we need to find the three sides of the triangles by taking the distance between the points.

Distance between two points is expressed as:

D = √(x2-x1)²+(y2-y1)²

For side |AB|:

A(-5, 4) and B(4, 1)

|AB| = √(4-(-5))²+(1-4)²

|AB| = √9²+3²

|AB| = √90

For side |BC|

B(4, 1), and C(1, -8)

|BC| =√(1-4)²+(-8-1)²

|BC| = √3²+9²

|BC| = √90

For side |AC|:

A(-5, 4) and C(1, -8).

|AC| = √(1-(-5))²+(-8-4)²

|AC| = √6²+12²

|AC| = √36+144

|AC| = √180

Based on the distances, it is seen that side AB and BC are equal which shows that two sides of the triangle are equal. A triangle that has two of its sides to be equal is known as an ISOSCELES TRIANGLE. Therefore the term that correctly describes the triangle is an isosceles triangle.

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4 years ago

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Phantasy [73]

Answer:

ΔABC≅ΔDEC by AAS

Step-by-step explanation:

You can use the AAS method of congruency.

Since you already have <BAC and <EDC congruent to eachother, and sides BC and EC congruent to each other, you only need that one remaining angle in between. <ACB can be proven congruent to <DCE by the Vertical Angles Theorem, and that gives you the AAS you need to prove that these two triangles are congruent

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5 0

3 years ago

Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).

I am Lyosha [343]

Take the homogeneous part and find the roots to the characteristic equation:

$y''+4y=0\implies r^2+4=0\implies r=\pm2i$

This means the characteristic solution is $y_c=C_1\cos2x+C_2\sin2x$ .

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form $y_p=ax\cos2x+bx\sin2x$ . Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With $y_1=\cos2x$ and $y_2=\sin2x$ , you're looking for a particular solution of the form $y_p=u_1y_1+u_2y_2$ . The functions $u_i$ satisfy

$u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$

where $W(y_1,y_2)$ is the Wronskian determinant of the two characteristic solutions.

$W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2$

So you have

$u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx$
$u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x$

$u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx$
$u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x$

So you end up with a solution

$u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

but since $\cos2x$ is already accounted for in the characteristic solution, the particular solution is then

$y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x$

so that the general solution is

$y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x$