Given:
The line passing through (-2,5) and (2,p) has a gradient of 
.
To find:
The value of p.
Solution:
If a line passes through two points, then the slope of the line is:
The line passing through (-2,5) and (2,p). So, the slope of the line is:
It is given that the gradient or slope of the line is .
On cross multiplication, we get
Divide both sides by 2.
Therefore, the value of p is 3.
Hello! For this you are just going to do this step by step, and remember to use PEMDAS. That is the order in which you should do the math. The abbreviation stands for Parenthesis, exponents, multiplication, division, addition and subtraction. So lets get started!
<span><span>(<span><span><span><span>(<span><span><span><span><span>15<span>y0</span></span><span>x<span>−3</span></span></span><span>z2</span></span>3</span><span>x3</span></span>)</span><span>(<span>y7</span>)</span></span><span>z1</span></span>2</span>)</span>2</span><span>=<span><span><span>254</span><span><span><span>x6</span><span>y14</span></span><span>z6</span></span></span><span>x6</span></span></span><span>=<span><span>254</span><span><span>y14</span><span>z6</span></span></span></span>
Hope this helps!! If you need any more help or further explanation just let me know!! :)
Answer:
The answer is "![\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%26-2%5C%5C%20-%5Cfrac%7B1%7D%7B2%7D%26%5Cfrac%7B7%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%7D)
".
Step-by-step explanation:
![\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D)
Solve the L.H.S part:
![\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D)
After calculating the L.H.S part compare the value with R.H.S:
![\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2B2c%26b%2B2d%5C%5C3a%2B4c%263b%2B4d%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C%2019%268%5Cend%7Barray%7D%5Cright%5D%7D%20%5C%5C%5C%5C)
In equation (i) multiply by 3 and subtract by equation (iii):
put the value of c in equation (i):
In equation (ii) multiply by 3 then subtract by equation (iv):
put the value of d in equation (iv):
The final answer is "".
