We have been given an expression
. We are asked to find the solution to our given expression expressed as scientific notation.
Let us simplify our given expression.
Using exponent property
, we will get:



Now to write our answer in scientific notation, we need our 1st multiple between 1 and 10. So we will rewrite our expression as:



Therefore, our required solution would be
.
"Quantitative" information is information that consists of numbers
and tells you the measure (quantity) of something.
The cost, mileage, and weight of the car are all quantitative items.
The model of the car may have numbers in it, but it doesn't tell
the measure of anything. It isn't quantitative data.
the answer to your question is 13 mm
Answer:
x1 = 6 + 2√6 or x2 = 6 - 2√6
Step-by-step explanation:
x² - 12x + 12 = 0
We need to use formula:
a² - 2ab + b² = (a - b)².
x² - 12x = - 12
x² - 2*6x + 6²- 6² = - 12
(x - 6)² - 36 = -12
(x - 6)² = - 12 + 36
(x -6)² = 24
(x - 6) = √24 or (x-6) = -√24
(x - 6) = 2√6 or (x-6) = - 2√6
x = 6 + 2√6 or x = 6 - 2√6
Answer: The determinant of the coefficient matrix is -15 and x = 3, y = 4, z = 1.
Step-by-step explanation: The given system of linear equations is :

We are given to find the determinant of the coefficient matrix and to find the values of x, y and z.
The determinant of the co-efficient matrix is given by

Now, from equations (ii) and (iii), we have

Substituting the value of y and z from equations (iv) and (v) in equation (i), we get

From equations (iv) and (v), we get

Thus, the determinant of the coefficient matrix is -15 and x = 3, y = 4, z = 1.