5

(6} Rumber 11 Robiem (a)v404S9uansanb ULD...

Question

(6} Rumber 11 Robiem (a)v404S9uansanb ULD

(6} Rumber 1 1 Robiem (a)v 404S 9 uansanb ULD



Answers

9. $$1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \dots$$

So in this question, we have to perform an operation. Are one entertained with our do. So it's very simple. We just have to rewrite the rose. So are you will become our one basically one minus three. Why do for on here? I have minus six nine and for so that's my answer.

To find these two matrices together. Here. First thing we need to do is check the dimensions will work. Okay? First matrix we've got is two rows three columns. That's a two by three. The second matrix is a three row one column matrix. These inner two dimensions have to match which they do. That means is a resulting matrix will be dimension the outer dimensions. Okay, So we're gonna have to buy one. Matrix is our answer just two spots here. Okay, we'll fill in the first spot. This is gonna be row one. Column one. Okay, so everything in row one thinks everything in column one negative one times six plus zero times negative four plus seven. Thanks one. Okay, this should give us negative six plus zero plus seven. That should give us a one in that first little spot right there. Okay, The second guy this is gonna be row two column one. Okay, so we're gonna take everything in row two times that same column right there. So three times six less negative five that was negative for plus two times one spent 18 plus 20 plus two should give us 40. Okay? So one in the first row, 40 in the second round, and that is the result of this modification.

Before this question, We just need Teoh simplify the expression, right? And so pretty much. We have this negative exponents whenever we have a negative. Excellent. We have to take the fraction of it, right? It's gonna be one over that entire expressions. What I mean by this is it would look something like this. You have won over, and then we go ahead and put this entire expression in the denominator of so we have is kind of how it looks. And this is just equivalent to 9/9 power of 5/2. Oops. 5/2. And another way to do this. Another way to right. This is not into the car of 1/2 to the power of five, right? Because whenever we have an exploding race for another exploding, all we have to do is multiply it. Now 90 Carbone has is just equivalent to the square root of nine, right? The power five school of Nice. Just three. So we have three that are five, and that's gonna be equivalent to 240

So I'm gonna solve this problem using penned ass. So I'm first gonna look for parentheses than exponents than multiplication and division left to right. In addition and subtraction, left to right is going to start by scanning through looking for parentheses. So I have a grouping symbol here and within that I have division and another set of parentheses till my first step in solving this problem is going to do my innermost parentheses first. So I've got six parentheses, 21 divided by three plus four is seven minus nine. So I go back to the start and I scan through looking for parentheses again. So even though there's one here, there's only one number inside. But this is a set of parentheses as well. So I'm gonna do this. Division six 21 divided by seven is three minus nine. I have two operations left. Have multiplication and subtraction, so I'm gonna do my multiplication first. Six times three is 18 18 minus nine is nine


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