Ask question

Ask question

All categories

Hunter-Best [27]

3 years ago

8

Find BC if segment AD is an altitude of triangle ABC PLEASE HELP

1 answer:

Vedmedyk [2.9K]3 years ago

3 0

Answer:

BC = 96

Step-by-step explanation:

Since AD is an altitude, then ∠ ADC = 90° , thus

4x + 10 = 90 ( subtract 10 from both sides )

4x = 80 ( divide both sides by 4 )

x = 20

Thus

BC = BD + DC

= 2x + 3x - 4

= 5x - 4

= 5(20) - 4 = 100 - 4 = 96

You might be interested in

The asymptote function f(x)=3^x+4 is y=

swat32

The asymptote of the function $f(x)=3^x$ is y = 0.

The asymptote of the function $f(x)=3^x+4$ is y = 0+4=4

Answer: y = 4

7 0

3 years ago

Child admission is $5.70 and adult admission is $9.50. Twice as many adult tickets ad child tickets were sold. for a total of $5

makkiz [27]

Let the number of child tickets sold be x, and number of adult tickets be y, then
5.70x + 9.50y = 543.40 . . . (1)
y = 2x . . . (2)

Putting (2) into (1), gives
5.7x + 9.5(2x) = 543.4
5.7x + 19x = 543.4
24.7x = 543.4
x = 543.4/24.7 = 22

Therefore, 22 child tickets were sold.

4 0

4 years ago

3/2-5/6-2/9<br><br> WILL GIVE BRAINLIEST THANKS AND 5 STARS FOR THE 1ST CORRECT ANSWER

Serga [27]

Answer:

4/9

Step-by-step explanation:

It’s basic math UwU

4 0

3 years ago

Read 2 more answers

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

3x - 9 + 4x<br> Please Help

Wewaii [24]

Answer:

Add like terms 7x - 9

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers