Answer:
A.29820
Step-by-step explanation:
39,594 - 9,774
29820
Answer:
Option A. y = x² + 7x + 10
Step-by-step explanation:
We'll begin calculating the roots of the equation from the graph.
The roots of the equation on the graph is where the curve passes through the x-axis.
The curve passes through the x-axis at –5 and –2
Next, we shall determine the equation. This can be obtained as follow:
x = –5 or x = –2
x + 5 = 0 or x + 2 = 0
(x + 5)(x + 2) = 0
Expand
x(x + 2) + 5(x + 2) = 0
x² + 2x + 5x + 10 = 0
x² + 7x + 10 = 0
y = x² + 7x + 10
Thus, the function that describes the graph is y = x² + 7x + 10
Answer:
b = 2
Step-by-step explanation:
Original Equation:
-12 + 3b - 1 = -5 - b
Add b to both sides
-12 + 4b - 1 = -5
Add 12 to both sides
4b - 1 = 7
Add 1 to both sides
4b = 8
Divide both sides by 4
b = 2
Hope this is what you need :)
Answer:
10.99
Step-by-step explanation:
where you see tan20 the 20 has a degree sign on top
Hello, let's note A the matrix, we need to find 
such that A= I, where I is the identity matrix, so the determinant is 0, giving us the characteristic equation as
We just need to solve this equation using the discriminant.
And then the eigenvalues are.
To find the basis, we have to solve the system of equations.
![A\lambda_1-\lambda_1 I=\left[\begin{array}{cc}3i&3\\-3&3i\end{array}\right] \\\\=3\left[\begin{array}{cc}i&1\\-1&i\end{array}\right] \\\\\text{For a vector (a,b), we need to find a and b such that.}\\\\\begin{cases}ai+b=0\\-a+bi=0\end{cases}\\\\\text{(1,-i) is a base of this space, as i-i=0 and -1-}i^2\text{=-1+1=0.}](https://tex.z-dn.net/?f=A%5Clambda_1-%5Clambda_1%20I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3i%263%5C%5C-3%263i%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%3D3%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Di%261%5C%5C-1%26i%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5Ctext%7BFor%20a%20vector%20%28a%2Cb%29%2C%20we%20need%20to%20find%20a%20and%20b%20such%20that.%7D%5C%5C%5C%5C%5Cbegin%7Bcases%7Dai%2Bb%3D0%5C%5C-a%2Bbi%3D0%5Cend%7Bcases%7D%5C%5C%5C%5C%5Ctext%7B%281%2C-i%29%20is%20a%20base%20of%20this%20space%2C%20as%20i-i%3D0%20and%20-1-%7Di%5E2%5Ctext%7B%3D-1%2B1%3D0.%7D)
![A\lambda_2-\lambda_2 I=\left[\begin{array}{cc}-3i&3\\-3&-3i\end{array}\right] \\\\=3\left[\begin{array}{cc}-i&1\\-1&-i\end{array}\right]\\\\\text{For a vector (a,b), we need to find a and b such that.}\\\\\begin{cases}-ai+b=0\\-a-bi=0\end{cases}\\\\\text{(1,i) is a base of this space as -i+i=0 and -1-i*i=0.}](https://tex.z-dn.net/?f=A%5Clambda_2-%5Clambda_2%20I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3i%263%5C%5C-3%26-3i%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%3D3%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-i%261%5C%5C-1%26-i%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Ctext%7BFor%20a%20vector%20%28a%2Cb%29%2C%20we%20need%20to%20find%20a%20and%20b%20such%20that.%7D%5C%5C%5C%5C%5Cbegin%7Bcases%7D-ai%2Bb%3D0%5C%5C-a-bi%3D0%5Cend%7Bcases%7D%5C%5C%5C%5C%5Ctext%7B%281%2Ci%29%20is%20a%20base%20of%20this%20space%20as%20-i%2Bi%3D0%20and%20-1-i%2Ai%3D0.%7D)
Thank you
