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Lct ((*) pointsand computc (hc ricmann Sum ovcr the intenral [',2] using Tollowind numbcr the Ielt Endpuits the subaltcrvals (Round Your unsners two decina pl...

Question

Lct ((*) pointsand computc (hc ricmann Sum ovcr the intenral [',2] using Tollowind numbcr the Ielt Endpuits the subaltcrvals (Round Your unsners two decina placesQnintconi; (Ni-cac caec cooen thc mycscntatiycUzc tj slbinteralzcqual IcnathUsc fIvc subintervals "qual erthUac ton qubintunnsoual Icgiathquese et Ine ared ol (he reqion under the oraph 0t / On the Interval [1 qqubre unils

Lct ((*) points and computc (hc ricmann Sum ovcr the intenral [',2] using Tollowind numbcr the Ielt Endpuits the subaltcrvals (Round Your unsners two decina places Qnintconi; (Ni- cac caec cooen thc mycscntatiyc Uzc tj slbinteralz cqual Icnath Usc fIvc subintervals "qual erth Uac ton qubintunns oual Icgiath quese et Ine ared ol (he reqion under the oraph 0t / On the Interval [1 qqubre unils



Answers

For the points $R(-4,2), S(1,4), T(3,-1),$ and $V(-7,-5),$ determine whether the Infin ars narallsh nernendiswlar, ar neither. $\overleftarrow{R S}$ and $\overleftarrow{V T}$

We're gonna figure out if rt and we are are parallel, perpendicular or neither. So we're gonna find out their slopes. So the soap of rt would be negative one minus two over three minus negative four. Which leaves you with negative three over positive seven, and that can't be reduced. And then to find the soap of the are we would do to minus and negative five over negative for minus negative seven. Which leaves me with a positive seven in a positive three. So this is up. They are opposite. Reciprocal. So that means that these two lines are perpendicular.

Hi everybody. So today we're going to be solving this system of equations and then we'll be plugging our A. B and C value into a quadratic problem formula. So we'll start by isolating see in this equation right here So we get four c. is equal to we'll just do it. This equation right here The four c. is equal to 25 -6 B -14 a. We're gonna raise it then we get C. Is equal to 25/4 6/4 -14 course and I'll just add me B. Okay now we're gonna plug that into our next two equations. So you want to go ahead and distribute the six out and when you do you'll end up with mhm. End up simplifying it down to five B plus 15. Okay. Us 75/2 is equal to 21. So then we'll just go in and subtract the 75/2 over. So trial in minutes sir, 75 house, forgive us- 16.5. So now we're going to go ahead and go back to our third and final equation and do the same thing when we distribute and simplify this down, you'll get 15 B plus 49. Hey plus 87 a half is equal to 33. We'll go ahead and subtract the 87 a half over. We'll get- 54.5. So now we're gonna use our this equation right here to solve for B. And they will plug into this equation here. Yeah 23. Subtract 15 a. over. Yeah. And divide by five. So we should get B. Is equal to negative three A minus 3.3 from here. We're going to go to this equation right here, Return to this uh that one The one above it. We'll go ahead and plug in our new value for B. When you simplify that down, you should get -45 A. Okay minus 49.5 Plus 49 A. Is equal to negative 54.5. Well combine like terms take it for A. Is equal to -5. So we get a. is equal to negative force. And from there we learned that B is equal to 0.45 and C. Is equal to 9.95. And from there we're just going to put this into the appropriate question. So why is equal to A. X squared plus B. X plus C. So we get Y. Is equal to negative 1.25 X squared us zero point 45 X. Yes. 9.95. We'll run scrolling and there's our problem and there you have it.

Hi everybody. So today we're going to be solving this system of equations and then we'll be plugging our A. B and C value into a quadratic problem formula. So we'll start by isolating see in this equation right here So we get four c. is equal to we'll just do it. This equation right here The four c. is equal to 25 -6 B -14 a. We're gonna raise it then we get C. Is equal to 25/4 6/4 -14 course and I'll just add me B. Okay now we're gonna plug that into our next two equations. So you want to go ahead and distribute the six out and when you do you'll end up with mhm. End up simplifying it down to five B plus 15. Okay. Us 75/2 is equal to 21. So then we'll just go in and subtract the 75/2 over. So trial in minutes sir, 75 house, forgive us- 16.5. So now we're going to go ahead and go back to our third and final equation and do the same thing when we distribute and simplify this down, you'll get 15 B plus 49. Hey plus 87 a half is equal to 33. We'll go ahead and subtract the 87 a half over. We'll get- 54.5. So now we're gonna use our this equation right here to solve for B. And they will plug into this equation here. Yeah 23. Subtract 15 a. over. Yeah. And divide by five. So we should get B. Is equal to negative three A minus 3.3 from here. We're going to go to this equation right here, Return to this uh that one The one above it. We'll go ahead and plug in our new value for B. When you simplify that down, you should get -45 A. Okay minus 49.5 Plus 49 A. Is equal to negative 54.5. Well combine like terms take it for A. Is equal to -5. So we get a. is equal to negative force. And from there we learned that B is equal to 0.45 and C. Is equal to 9.95. And from there we're just going to put this into the appropriate question. So why is equal to A. X squared plus B. X plus C. So we get Y. Is equal to negative 1.25 X squared us zero point 45 X. Yes. 9.95. We'll run scrolling and there's our problem and there you have it.

So we wouldn't know if RV and str parallel perpendicular or neither. So we have RV. We're gonna first find our slopes. So that's negative. Five minus two over negative seven minus negative for. And I will leave me with negative seven and a negative three, Which leaves me with just 7/3. And then for S t. We have a negative one minus four and then three minus one, which is with negative five over to which does not simplify. So these are neither the same nor the opposite. Reciprocal. So these two lines are neither.


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