Answer:
![\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%265%5C%5C13%26-13%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
If you have two matrices:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}a+e&b+f\\c+g&d+h\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2Be%26b%2Bf%5C%5Cc%2Bg%26d%2Bh%5Cend%7Barray%7D%5Cright%5D)
We have:
![A=\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]\\and\\B=\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D)
And we need to express as a single matrix:
![A+B=\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}6+(-2)&-3+8\\10+3&-1+(-12)\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}6-2&5\\13&-1-12\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]](https://tex.z-dn.net/?f=A%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%2B%28-2%29%26-3%2B8%5C%5C10%2B3%26-1%2B%28-12%29%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6-2%265%5C%5C13%26-1-12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%265%5C%5C13%26-13%5Cend%7Barray%7D%5Cright%5D)
The answer is:
It is expressed as a single matrix.
Answer: 6/121
Step-by-step explanation:
P(not blue)= P(red) × P(green)
=6/11 × 1/11
6/121
Answer:
a) Fx=-5N Fy=-5*sqrt(3) N b) Fx= 5*sqrt(3) N Fy=-5N
c) Fx=-5*sqrt(2) N Fy=-5*sqrt(2) N
Step-by-step explanation:
The arrow's F ( weight) component on axle x is Fx= F*sinA and on axle y is
Fy=F*cosA
a) The x component and y component both are opposite directed to axle x and axle y accordingly. So both components are negative.
So Fx = - 10*sin(30)= -5 N Fy= -10*cos(30)= -10*sqrt(3)/2= -5*sqrt (3) N
b) Now the x component is co directed to axle x , and y component is opposite directed to axle y.
So x component is positive and y components is negative
So Fx = 10*sin(60)= 5*sqrt(3) N Fy= -10*cos(60)= -10*1/2= -5 N
c)The x component and y component both are opposite directed to axle x and axle y accordingly. So both components are negative.
So Fx = - 10*sin(45)= -5*sqrt(2) N
Fy= -10*cos(45)= -10*sqrt(2)/2= -5*sqrt (2) N
Answer:
This skill will come in handy when working with word problems or real life situations.
Step-by-step explanation:
