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3 years ago

Given: BD bisects ABC. Auxiliary EA is drawn such that AE || BD. Auxiliary BE is an extension of BC

Mathematics

2 answers:

AnnZ [28]3 years ago

7 0

The required proof is given in the table below:

$\begin{tabular}{|p{4cm}|p{6cm}|} 
 Statement & Reason \\ [1ex] 
1. $\overline{BD}$ bisects $\angle ABC$ & 1. Given \\
2. \angle DBC\cong\angle ABD & 2. De(finition of angle bisector \\ 
3. $\overline{AE}$||$\overline{BD}$ & 3. Given \\ 
4. \angle AEB\cong\angle DBC & 4. Corresponding angles \\
5. \angle AEB\cong\angle ABD & 5. Transitive property of equality \\ 
6. \angle ABD\cong\angle BAE & 6. Alternate angles
\end{tabular}$
$\begin{tabular}{|p{4cm}|p{6cm}|}
7. \angle AEB\cong\angle BAE & 7. Transitive property of equality \\
8. \overline{EB}\cong\overline{AB} & 8. From 7. $\Delta ABE$ is isosceles \\
9. EB = AB & 9. De(finition of congruence \\ 10. $\frac{AD}{DC}=\frac{EB}{BC}$ & 10. Triangle proportionality theorem \\
11. $\frac{AD}{DC}=\frac{AB}{BC}$ & 11. Substitution Property of equality \\[1ex] 
\end{tabular}
$

34kurt3 years ago

7 0

Answer:

Reasons 2 and following

2. Definition of angle bisector

3. Given

4. They are corresponding angles

5. Transitive property of equality

6. Alternate angles are congruent

7. Transitive property of equality

8. From 7 ΔABE is isosceles

9. Definition of isosceles triangle

10. Triangle proportionality theorem

11. Substituting 9 in 10

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$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

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$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

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and our differential equation (*) gets transformed in the algebraic equation

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$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

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valentina_108 [34]

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