Answer:
<h2>1. x = 4</h2><h2>2. x = 20</h2>
Step-by-step explanation:
1.
ΔABC and ΔAJK are similar (AA). Therefore the sides are in proportion:
We have:
AC = 1 + 4 = 5
AJ = 1
AB = 1 + x
AK = 1
Substitute:
<em>subtract 1 from both sides</em>
2.
ΔVUT and ΔVMN are similar (AA). Therefore the sides are in proportion:
We hve:
VU = x + 8
VM = x
VT = 49
VN = 49 - 14 = 35
Substitute:
<em>cross multiply</em>
<em>use the distributive property a(c + b) = ab + ac</em>
<em>subtract 35x from both sides</em>
<em>divide both sides by 14</em>
Answer:
Step-by-step explanation:
We have given that melissa used 4 cups of sugar to bake 54 cookies.
That means 1 cup of sugar will be used to bake 
by unitary method
We have to find the ratio of cups to the cookies
We are using 1 cup of sugar
And cookies will be 
Answer:
5:24
Step-by-step explanation:
We're provided with the number of rebounds as 90 while the points are 432. Expressing them into ratio of rebounds to steals we have
90:432
Simplification:
Dividing both sides by 2 we obtain
45:216
Dividing both sides of the above ratio by 3 we obtain
15:72
Dividing both sides of the above ratio further by 3 we obtain
5:24
Therefore, rhe simplified ratio of rebounds ro steals is 5:24
Answer:
We have the equation
![c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]](https://tex.z-dn.net/?f=c_1%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_2%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_3%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_4%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Then, the augmented matrix of the system is
![\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D0%260%260%268%5C%5C0%260%264%264%5C%5C0%263%263%263%5C%5C1%261%261%261%5Cend%7Barray%7D%5Cright%5D)
We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:
![\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%261%5C%5C0%263%263%263%5C%5C0%260%264%264%5C%5C0%260%260%268%5Cend%7Barray%7D%5Cright%5D)
This matrix is in echelon form. Then, now we apply backward substitution:
1.
2.
3.
4.
Then the system has unique solution that is 
and this imply that the vectors 
are linear independent.
Answer:
1/49
<em>Step by step explanation:</em>
