Question

Will fan, medal, and write testimony! Please work with me to solve this problem!

Answer 1

@Directrix Could you help me with this?

Answer 2

Quad RXTZ is a trapezoid with median XY.

The median of a trapezoid has measure 1/2 of the sum of the lengths of the two bases of the trapezoid.

43 = (1/2) * (x^2 + 7 + 2x^2 + 4)

Solve that for x: @bobbyjack1200

Answer 3

@Directrix 43= (1/2) * 3+x^2?

Answer 4

That does not look right.

Do this:

Simplify this: (x^2 + 7 + 2x^2 + 4) = ?

Answer 5

@Directrix 3x2+11?

Answer 6

Now how do I find the segments?

Answer 7

@pooja195

Answer 8

@TheSmartOne

Answer 9

I know X = 3x^2+11 but how do I find the segment

Answer 10

@TheSmartOne Are you able to help me?

Answer 11

Sorry, maybe @Hero @Nnesha @zepdrix can help you :)

Answer 12

Okay thanks for trying :) @TheSmartOne

Answer 13

:)

Answer 14

@thomaster @Whitemonsterbunny17 What about you guys? Can you please help?

Answer 15

@IrishBoy123 @Preetha Could you help me?

Answer 16

@Nnesha Are you able to help?

Answer 17

Idk but I'll try to find people to help. :/

Answer 18

If only OwlCoffee was on. I'm sure he would be able to help. :|

Answer 19

@Willie579 Thank you I apprieciate it

Answer 20

Wow this is hard... ;-;

Answer 21

x= what ??
look at the information provided by directrix solve for x

Answer 22

@Nnesha I did. I already stated above that I know X = 3x^2+11 but how do I find the segments?

Answer 23

no you still have to solve for x
it should be x= number

Answer 24

you got the quadratic equation do you know quadratic formula ?? use that to solve for x

Answer 25

@Nnesha How do I solve beyond 3x^2+11? We haven't learned the quadratic equation?

Answer 26

well ... the equation is \[\rm \large 43=\frac{ 1 }{ 2}(\color{Red}{x^2+2x^2+7+4})\]
red part equal to 3x^2+!1

Answer 27

3x^2+11 **solve the rest of it

Answer 28

@Nnesha Yes so to do the quadratic equation, would I multiple 3x^2+11 by 1/2?

Answer 29

yes multiply by 1/2

Answer 30

@Nnesha All I get is 14.5?

Answer 31

Or 29/2?

Answer 32

actually get rid of the 2 from right side

Answer 33

\[43 =\frac{ (3x^2+11) }{ 2 }\] how would you cancel out 2 ?

Answer 34

Multiply it by itself?

Answer 35

multiply both sides by 2

Answer 36

Okay so then I would have 6x^2+22?

Answer 37

we are multiplying by 2 to cancel out the 2 from right side \[43 *2=(\frac{ 3x^2+11 }{ 2 })*2\]

Answer 38

So we would do 43 times 2 and 11 times 2? Or 3x^2 times 2 and 11 times 2?

Answer 39

no just cancel out the 2's
\[43 *2=(\frac{ 3x^2+11 }{ \cancel{2} })*\cancel{2}\]
that's the reason we were multiplying 2 boht sides

Answer 40

Ohhh so what about the 2 over by the 43 though??

Answer 41

that's multiplication 43 times 2

Answer 42

So then we would have 86=3x^2+11?

Answer 43

@Nnesha

Answer 44

Im confused.....

Answer 45

@Directrix Am I doing this right?

Answer 46

The original equation is this:

43 = (1/2) * (x^2 + 7 + 2x^2 + 4)

43 = (1/2) * (3x^2 + 11) <------------

Is this where you are now? @bobbyjack1200

Answer 47

Oh, I see that you are here:

>>>So then we would have 86=3x^2+11? after multiplying by 2

Answer 48

\(\color{blue}{\text{Originally Posted by}}\) @bobbyjack1200
So then we would have 86=3x^2+11?
\(\color{blue}{\text{End of Quote}}\)
that's correct
sorry i was looking n other tab and i have to go now Directrix is online so he/she will help u

Answer 49

43 = (1/2) * (3x^2 + 11)

86 = 3x^2 + 11 is correct so far

Subtract 11 from both sides. @bobbyjack1200

Answer 50

@Directrix So then id have 75=3x^2?

Answer 51

86 = 3x^2 + 11
-11 - 11
----------------------

Answer 52

> 75=3x^2

Yes.

Now, divide both sides by 3. Post what you get.

Answer 53

So then Id have 25=x^2?

Answer 54

I dont think I can simplify past this, can I?

Answer 55

Yes, you can.

x^2 = 25

x^2 - 25 = 0

(x + 5) * (x - 5) = 0

x = -5 or x = 5 Two values for x.

Answer 56

Find RX.

RX = x^2 + 7

RX = (5)^2 + 7
RX = ?

RX = (-5)^2 + 7
RX = ?

Answer 57

@Directrix Are you sure thats what they are asking for in this problem?

Answer 58

Two values I mean?

Answer 59

Im confused? Didnt we just solve for RX?

Answer 60

Oh Im sorry, I see. we just solved for X

Answer 61

Now we solve for RX

Answer 62

We solved for x.

x and RX are not the same.

Read Part One and see what you think they are solving for.

I'm looking at RX and TZ still left to find if you would just do it.

I set up RX for you already.

Answer 63

For RX we have 32 and -18, correct?

Answer 64

and for TZ we would have 54 and -46?

Answer 65

Am I right?

Answer 66

@Directrix

Answer 67

Now for pt.2 What do you suggest would be the best way to find the angles they want me to find?

Answer 68

RX = (5)^2 + 7

RX = 25 + 7

RX = 32

---

RX = (-5)^2 + 7 Note: Negative 5 times negative 5 = + 25

RX = 25 + 7

RX = 32


RX is 32 <----------

Answer 69

>>For RX we have 32 and -18, correct?

No.

Answer 70

@Directrix Im sorry I had fogotten a Negative and negative make a positive

Answer 71

TZ = 2*x^2 + 4

TZ = 2 * (5^2) + 4

TZ = 54 <------------------

Same TZ value if x = -5.

Answer 72

You have been on this problem for 4 hours.

Let's finish it and move on.

Answer 73

Part One is complete.

Answer 74

@Directrix Okay thank you for all your patience. I really appreciate it :)

Answer 75

<XYS is also 30 because:
if two parallel lines are cut by a transversal, corresponding angles are congruent.