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3 years ago

5

Triangle ABC has a horizontal sido AC, and AB in perpendicular to BC. A vertical line is drawn from a point labeled Don AC to th

e point 8.
Which of the following statements are true? Select the two correct answers.

1 answer:

SashulF [63]3 years ago

5 0

Answer:

Option (3)

Step-by-step explanation:

Given:

Horizontal side of a triangle ABC is side AC.

AB and BC are the perpendicular lines.

BD is perpendicular to side AC.

Solution:

Option (1).

Slopes of AB and BC are reciprocal.

False.

Option (2)

Angle DAB and BCD are equal.

False.

Option (3).

Since, sides AB and BC are perpendicular lines,

Slopes of AB and BC will be negative reciprocals.

True.

Option (4).

Triangles ABD and BCD are similar.

False.

Therefore, Option (3) will be the answer.

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The coordinates of the midpoint of GH¯¯¯¯¯¯¯¯are M(8, 0) and the coordinates of one endpoint are H(−2, 9).

mixer [17]

Answer:

G(18;-9).

Step-by-step explanation:

the x-coordinate of the point G is:

$X_G=2*X_M-X_H=16+2=18;$

the y-coordinate of the point G is:

$Y_G=2*Y_M-Y_H=0-9=-9.$

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3 years ago

The arithmetic sequence ai is defined by the formula: a 1 =15 a i =a i-1 -7 Find the sum of the first 660 terms in the sequence.

lara31 [8.8K]

Answer:

The sum of the first 660 terms in the sequence=−1512390

Step-by-step explanation:

We are given that an arithmetic sequence

$a_1=15$

$a_{i}=a_{i-1}-7$

We have to find the sum of first 660 terms

Substitute i=2

$a_2=a_1-7=15-7=8$

Substitute i=3

$a_3=a_2-7=8-7=1$

Now, the common difference

$d=a_2-a_1=8-15=-7$

Now, sum of n terms of an A.P

$S_n=\frac{n}{2}(2a+(n-1)d)$

Substitute n=660, d=-7, a=15

$S_{660}=\frac{660}{2}(30+(660-1)(-7))$

$S_{660}=-1512390$

Hence, the sum of the first 660 terms in the sequence=−1512390

6 0

3 years ago

Determine the equation of the line connecting the points (0,-1) and (2,3)

Nutka1998 [239]

Y = 2x - 1 is the equation

7 0

4 years ago

Y=1/4 x y=4x-5 parallel perpendicular neither

ziro4ka [17]

Answer:

Neither.

Step-by-step explanation:

The slope for the first one is 1/4 and the slope for the second one is 4. These are reciprocals, but not opposite reciprocals, meaning they are not perpendicular. An opposite reciprocal means the fraction is flipped (ex. 3/1 to 1/3) and the sign is changed. The line parallel to y=1/4x would be y= -4x.

3 0

3 years ago

Consider the matrix A. A = 1 0 1 1 0 0 0 0 0 Find the characteristic polynomial for the matrix A. (Write your answer in terms of

dusya [7]

Answer with Step-by-step explanation:

We are given that a matrix

$A=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

a.We have to find characteristic polynomial in terms of A

We know that characteristic equation of given matrix $\mid{A-\lambda I}\mid=0$

Where I is identity matrix of the order of given matrix

I= $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

Substitute the values then, we get

$\begin{vmatrix}1-\lambda&0&1\\1&-\lambda&0\\0&0&-\lambda\end{vmatrix}=0$

$(1-\lambda)(\lamda^2)-0+0=0$

$\lambda^2-\lambda^3=0$

$\lambda^3-\lambda^2=0$

Hence, characteristic polynomial = $\lambda^3-\lambda^2=0$

b.We have to find the eigen value for given matrix

$\lambda^2(1-\lambda)=0$

Then , we get $\lambda=0,0,1-\lambda=0$

$\lambda=1$

Hence, real eigen values of for the matrix are 0,0 and 1.

c.Eigen space corresponding to eigen value 1 is the null space of matrix $A-I$

$E_1=N(A-I)$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&-1\end{array}\right]$

Apply $R_1\rightarrow R_1+R_3$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]$

Now,(A-I)x=0[/tex]

Substitute the values then we get

$\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

Then , we get $x_3=0$

And $x_1-x_2=0$

$x_1=x_2$

Null space N(A-I) consist of vectors

x= $\left[\begin{array}{ccc}x_1\\x_1\\0\end{array}\right]$

For any scalar $x_1$

$x=x_1\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

$E_1=N(A-I)=Span(\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Hence, the basis of eigen vector corresponding to eigen value 1 is given by

$\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Eigen space corresponding to 0 eigen value

$N(A-0I)=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

$(A-0I)x=0$

$\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

$\left[\begin{array}{ccc}x_1+x_3\\x_1\\0\end{array}\right]=0$

Then, $x_1+x_3=0$

$x_1=0$

Substitute $x_1=0$

Then, we get $x_3=0$

Therefore, the null space consist of vectors

$x=x_2=x_2\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

Therefore, the basis of eigen space corresponding to eigen value 0 is given by

$\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

5 0

3 years ago