Answer:
(x-9)(x+4)
Step-by-step explanation:
When factoring, you use the formula ab ÷ c to find what you can multiply to get (in this case) -36 that you can add to get -5. The answer is -9 and 4.
Answer:
Step-by-step explanation:
Since the picture isn't given. I have no idea what the radius or diameter is and as such, have no definite answer for you.
However, from the question, if he's to run round the circular track once, then he must have run a total of
2πr meters, with r being the radius.
Now, the question says that he ran round the track 10 times, this means that whatever value we get there, for 1 trial, we multiply it by 10, to get the value for 10 trials. Essentially,
10 * 2πr
The value gotten is the needed answer. All you have to do is substitute r for the value you have in your diagram. We already know that π is 3.142
Answer:
Adults = x
Students = y = x + 40. 40 more student than adult
Adult = $8, Student = $6
Total for Adults = 8*x = 8x
Total for students = 6*y = 6y
Total = $1920
8x + 6y = 1920
Hence system of equations are:
y = x + 40
8x + 6y = 1920
Answer:
The image of ![\left[\begin{array}{c}4&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D)
through T is ![\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We know that 
→ is a linear transformation that maps 
into 
⇒
And also maps 
into 
⇒
We need to find the image of the vector
We know that exists a matrix A from 
(because of how T was defined) such that :
for all x ∈
We can find the matrix A by applying T to a base of the domain ().
Notice that we have that data :
{
}
Being 
the cannonic base of
The following step is to put the images from the vectors of the base into the columns of the new matrix A :
![T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%260%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%265%5Cend%7Barray%7D%5Cright%5D)
(Data of the problem)
![T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%261%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-2%267%5Cend%7Barray%7D%5Cright%5D)
(Data of the problem)
Writing the matrix A :
![A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C5%267%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now with the matrix A we can find the image of ![\left[\begin{array}{c}4&-4\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5C%5C%5Cend%7Barray%7D%5Cright%5D)
such as :
⇒
![T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]](https://tex.z-dn.net/?f=T%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C5%267%5C%5C%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%26-4%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D24%26-8%5Cend%7Barray%7D%5Cright%5D)
We found out that the image of through T is the vector
