Answer:
B
Step-by-step explanation:
x=√7
squaring
x²=7
Answer:
AB: size 4 x 4
BA: size 2 x 2
Step-by-step explanation:
When we want to multiply two matrices, we need first to check if the number of columns of the first matrix is equal the number of rows of the second matrix. Then, the resulting matrix would have the number of rows equal the number of rows of the first matrix, and the number of columns equal the number of columns of the second matrix.
That is, if we have a matrix A with dimensions a x b and a matrix B with dimensions c x d, to calculate the product AB we must have b = c, and the resulting product would have dimensions a x d.
Finally, if A is 4 x 2 and B is 2 x 4, we would have:
AB = (4 x 2) (2 x 4)
2 = 2 -> product is possible.
Dimensions of the result: 4 x 4
BA = (2 x 4) (4 x 2)
4 = 4 -> product is possible.
Dimensions of the result: 2 x 2
Answer:
This triangle is A, Scalene and E, Obtuse.
Step-by-step explanation:
- it is scalene because the lengths of all of its sides are different
- it is obtuse because it has an angle bigger than 90 degrees.
Hope this helps!
Notation
I imagine that the expression you are asked to work with is:

When you use a keyboard it is customary to use "^" to denote an exponent is coming so you could have written: 3x^3y+15xy-9x^2y-45y just to be clear.
PART A
To factor out the GCF we are looking for the greatest factor among the terms. Looking at the coefficients (the numbers) the largest number they can all be divided by is 3 so we will pull out a 3. Notice also that each term has a y in it so we can pull out that.
This gives us:

To factor is to write as a product (something times something else). It undoes multiplication so in this case if you take what we got and multiplied it back you should get the expression we started with.
PART B
Start with the answer in part A. Namely,

. For now let's focus only on what is in the parenthesis. We have four terms so let's take them two at a time. I am separating the expression in two using square brackets.
![[( x^{3}+5x)]-[3 x^{2} -15]](https://tex.z-dn.net/?f=%5B%28%20x%5E%7B3%7D%2B5x%29%5D-%5B3%20x%5E%7B2%7D%20-15%5D)
Let's next factor what is in each bracket:
![[( x^{3}+5x)]-[3 x^{2} -15] = [x( x^{2} +5)]-[3( x^{2} +5)]](https://tex.z-dn.net/?f=%5B%28%20x%5E%7B3%7D%2B5x%29%5D-%5B3%20x%5E%7B2%7D%20-15%5D%20%3D%20%5Bx%28%20x%5E%7B2%7D%20%2B5%29%5D-%5B3%28%20x%5E%7B2%7D%20%2B5%29%5D)
Notice that both brackets have the same expression in them so now we factor that out:
![[x( x^{2} +5)]-[3( x^{2} +5)] = (x-3)( x^{2} +5)](https://tex.z-dn.net/?f=%20%5Bx%28%20x%5E%7B2%7D%20%2B5%29%5D-%5B3%28%20x%5E%7B2%7D%20%2B5%29%5D%20%3D%20%28x-3%29%28%20x%5E%7B2%7D%20%2B5%29)
Our original expression (the one we started the problem with) had a 3y we already pulled out. We need to include that in the completely factored expression. Doing so we get: