The coefficient matrix is build with its rows representing each equation, and its columns representing each variable.
So, you may write the matrix as
![\left[\begin{array}{cc}\text{x-coefficient, 1st equation}&\text{y-coefficient, 1st equation}\\\text{x-coefficient, 2nd equation}&\text{y-coefficient, 2nd equation} \end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Ctext%7Bx-coefficient%2C%201st%20equation%7D%26%5Ctext%7By-coefficient%2C%201st%20equation%7D%5C%5C%5Ctext%7Bx-coefficient%2C%202nd%20equation%7D%26%5Ctext%7By-coefficient%2C%202nd%20equation%7D%20%5Cend%7Barray%7D%5Cright%5D%20%20)
which means
![\left[\begin{array}{cc}4&-3\\8&-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-3%5C%5C8%26-3%5Cend%7Barray%7D%5Cright%5D%20%20)
The determinant is computed subtracting diagonals:
![\left | \left[ \begin{array}{cc}a&b\\c&d\end{array}\right]\right | = ad-bc](https://tex.z-dn.net/?f=%20%5Cleft%20%7C%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5Cright%20%7C%20%3D%20ad-bc%20)
So, we have
![\left | \left[\begin{array}{cc}4&-3\\8&-3\end{array}\right] \right | = 4(-3) - 8(-3) = -4(-3) = 12](https://tex.z-dn.net/?f=%20%5Cleft%20%7C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-3%5C%5C8%26-3%5Cend%7Barray%7D%5Cright%5D%20%5Cright%20%7C%20%3D%204%28-3%29%20-%208%28-3%29%20%3D%20-4%28-3%29%20%3D%2012%20%20)
So it tells us that g(3) = -5 and g'(x) = x^2 + 7.
So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))
now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16
Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84
So we have our linear approximation for the two.
If you wanted to, you could check your answer by finding g(x). Since you know g'(x), take the antiderivative and we will get
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35
So just to check our linear approximations use that to find g(2.99) and g(3.01)
g(2.99) = -5.1597
g(3.01) = -4.8397
So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer. Not a bad method if you ever need to use it.
Answer:
8
Step-by-step explanation:
Formula for height of triangle
A÷BX2
In this case, 48÷12=4×2=8
Hope you understand ☺️
Your answer is E. $25.
First let under 12 = u, over 12 = o, and adults = a.
We can now write the equations:
2u + 3a + 3o = 174
4u + 2a = 122
a + o = 46
Because we know that a + o = 46, and 3a + 3o is in the first equation, we can multiply 46 by 3 to get what 3a + 3o equals. This makes 138.
Now we can substitute 138 into the first equation to get 2u + 138 = 174
2u = 36
u = 18
Now that we know what u equals, we can substitute it in to the second equation to get:
4(18) + 2a = 122
72 + 2a = 122
2a = 50
a = $25
I hope this helps! Let me know if you have any questions :)
Put the value of x = -1 to the equation of a function f(x):
