According to the secant-tangent theorem, we have the following expression:

Now, we solve for <em>x</em>.

Then, we use the quadratic formula:
![x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Where a = 1, b = 6, and c = -315.
![\begin{gathered} x_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_{1,2}=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_{1,2}=\frac{-6\pm36}{2} \\ x_1=\frac{-6+36}{2}=\frac{30}{2}=15 \\ x_2=\frac{-6-36}{2}=\frac{-42}{2}=-21 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B6%5E2-4%5Ccdot1%5Ccdot%28-315%29%7D%7D%7B2%5Ccdot1%7D%20%5C%5C%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B36%2B1260%7D%7D%7B2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B1296%7D%7D%7B2%7D%20%5C%5C%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm36%7D%7B2%7D%20%5C%5C%20x_1%3D%5Cfrac%7B-6%2B36%7D%7B2%7D%3D%5Cfrac%7B30%7D%7B2%7D%3D15%20%5C%5C%20x_2%3D%5Cfrac%7B-6-36%7D%7B2%7D%3D%5Cfrac%7B-42%7D%7B2%7D%3D-21%20%5Cend%7Bgathered%7D)
<h2>Hence, the answer is 15 because lengths can't be negative.</h2>
Answer:
Infinitely Many Solutions
Step-by-step explanation:
There can be infinite solutions to this problem
Here are some examples
3 4
6 8
12 16
24 32
48 64
96 128
To get a solution, just multiply both numbers by the same term. I multiplied by 2 each.
Hope this helps :)
-jp524
The base should be 4.
Area=1/2* b *h
4*7=28/2=14.
Answer:
The value of AB is
and it's not possible to multiply BA.
Step-by-step explanation:
Consider the provided matrices.
, ![B=\left[\begin{array}{ccc}3\\5\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
Two matrices can be multiplied if and only if first matrix has an order m × n and second matrix has an order n × v.
Multiply AB
Matrix A has order 2 × 2 and matrix B has order 2 × 1. So according to rule we can multiply both the matrix as shown:
![AB=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right] \left[\begin{array}{ccc}3\\5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C2%261%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}2\times 3+3\times 5\\2\times 3+1\times 5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5Ctimes%203%2B3%5Ctimes%205%5C%5C2%5Ctimes%203%2B1%5Ctimes%205%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}6+15\\6+5\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%2B15%5C%5C6%2B5%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}21\\11\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C11%5Cend%7Barray%7D%5Cright%5D)
Hence, the value of AB is ![\left[\begin{array}{ccc}21\\11\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D21%5C%5C11%5Cend%7Barray%7D%5Cright%5D)
Now calculate the value of BA as shown:
Multiply BA
Matrix B has order 2 × 1 and matrix A has order 2 × 2. So according to rule we cannot multiply both the matrix.
We can multiply two matrix if first matrix has an order m × n and second matrix has an order n × v.
That means number of column of first matrix should be equal to the number of rows of second matrix.
Hence, it's not possible to multiply BA.