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Question % [$ pts] Let A =Find all possibk values oftlutnorm(a,4) =15...

Question

Question % [$ pts] Let A =Find all possibk values oftlutnorm(a,4) =15

Question % [$ pts] Let A = Find all possibk values of tlut norm(a,4) =15



Answers

Evaluate the expression for the given values of the variables. $$7 a, \text { for } a=465$$

So if we're asked to find the A B, we know that air is 465 on B is equal to 32 and we simply just asked to multiply the two together, which will give us an answer off 14,000 880.

We have a function f of X equals X squared plus 14 X plus 50. We need to find all of the values of X so that the function is equal to five. So first we're going to do this algebraic lee, and then we're going to see what this looks like on a graph. To start, we're going to replace R F of X from our function with five. So five equals X squared plus 14 x plus 50. In order to solve a polynomial equation, it needs to be said equal dizzy room. So let's subtract five from both sides. Zero equals X squared plus 14 x plus 45. Since our lead coefficient is one, this will be relatively easy to factor. We know our first term of each factor is going to be X. We also know that both terms or both factors need to be sums because our product is positive and our some is positive. All that's left is find the factor pair of 45 that adds to give us 14 possibilities 15 and three or nine and five, nine and five are the pair that will add to 14. So let's use those we have the equation now zero equals X plus nine times X plus five. We can apply our principles of zero products and set each of these factors equal to zero. Next plus nine equals zero and X plus five equals zero to give us the two values that make our equation true. For our first factor subject nine from both sides and we have a solution of X equals negative nine for our second factor, X plus five equals zero. Subtract five from both sides and our second solution is X equals negative five. Now keep in mind these are not the zeros of our function, but they are the values of X that make our function equal to five. Graphically hopping over to Dismas, we can type in our function. F of X is equal to x squared plus 14 x That's 50 so we can see that our function isn't crossing the x axis. We are not looking for zeros. Instead, we want the values of our function or values of X that make our function equal to five. So five would be right around here to make this easier to visualize weaken graph a second function that is equal to five. Now we can see where our two functions intersect. This is when our first function X squared plus 14 x plus 50 and f of X equals five. Those points of intersection are when they're equal. So we see here X equals negative nine and X equals negative five. Back to our algebraic method. We have the solutions X equals negative nine and X equals negative five. They both make her function equal five.

If we had a function and we didn't know a sub N. But we did know that F of three was -150. Well then that means what if X. Is three. So let's start with plugging three into what we do know for X. Three to the fourth minus three times three Squared -4. Three times The 3 to the 4th is 81 -3 times three squared is nine. That's 81 -27 -4, Which is equal to 50. But we know that f of three is -150. So if all of this is going to be 50, then we know to get a negative 1 50 A sub N must be equal to negative three.

So no, we have effects is equals to 15 minus three x so as to be so as and fix has to be zero. So now for that they will just equals zero is because too 15 minus three x So if we equate, it would be three. X is because to 15. So therefore X would be fight because we divided three on both sides. So therefore excess because to fight so therefore X has to be five for effects is equals to zero.


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