Answer:
Five more than the square of a number= 5 + x²
Five more than twice a number = 5 + 2x
Five less than the product of 3 and a number = 3x - 5
Five less the product of 3 and a number = 5 -3x
Twice the sum of a number and 5 = 2(x + 5)
The sum of twice a number and 5 = 2x + 5
The product of a cube of a number and 5= 5x³
The cube of the product of 5 and a number= (5x)³
I think the answer is 61 dont quote me on it im awful at math
Answer:
no, they are not functions
Step-by-step explanation:
this is because a single domain has multiple range.
i.e. domain 2 has ranges 2 and 4
Step-by-step explanation:
For two points (x₁, y₁) and (x₂, y₂), the distance between them is:
d² = (x₁ − x₂)² + (y₁ − y₂)²
The order of points 1 and 2 don't matter. You can switch it:
d² = (x₂ − x₁)² + (y₂ − y₁)²
This is basically the Pythagorean theorem for a coordinate system.
Let's do an example. If you have points (1, 2) and (4, 6), then the distance between them is:
d² = (4 − 1)² + (6 − 2)²
d² = 3² + 4²
d² = 9 + 16
d² = 25
d = 5
If you have points with negative coordinates, remember that subtracting a negative is the same as adding a positive.
For example, the distance between (-1, -2) and (4, 10) is:
d² = (4 − (-1))² + (10 − (-2))²
d² = (4 + 1)² + (10 + 2)²
d² = 5² + 12²
d² = 25 + 144
d² = 169
d = 13
Answer:
Step-by-step explanation:
Given
![A = \left[\begin{array}{cc}-2&6\\3&5\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%266%5C%5C3%265%5Cend%7Barray%7D%5Cright%5D)
Required
Determine the determinant
For a two by two matrix, A such that:
![A = \left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
The determinant |A| is:
So, in
The determinant is:
