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Nice, CU  !!!!

This uses the Exterior Angle Theorem

It says that

measure of DBA  = measure of BAC + measure of  ACB

124  = 95  + measure of ACB    subtract 95 from both sides

29°  =   measure of ACB

You did question 1 wrong.

Wrong: 

       0    8

  - ( -3   4)

------------------

      -3   4

Right:

       0    8

  - ( -3   4)

------------------

      3   4

Here my hint:

To find the answers, first find situations for the answer choices where it is false. This will let you eliminate and pinpoint down answers.

You also may need to touch up on some theorems for example SAS or AAS

Use your imagination!

If the blue triangle is the original triangle, and the red triangle is the new triangle. Notice how the red is placed where it shares two vertices with.

There is no other way to place it or you way change the amount of vertices shared and the triangle's sides are all unequal

There is only 3 ways

I will give only hints as this problem only requires basic addition and subtraction

The sum of the angles of Triangle ABC add up to 180

Line segment DBC is 180 degrees, and if DBA is 124, then what must be ABC?

Dont be intimidated! You don't even need to figure out the exact angle measures of ABC and ACB! Just find the sum of their angle measures. (You may need to scroll horizontally)

\(\text{Triangle ABC's angles add up to 180. And since angle ABC and ACB are bisected. We can assign them values }\frac{1}{2}x\text{ and }\frac{1}{2}y\)

\(\text{We draw triangle PBC as shown in the image in red}\)

\(\text{Due to the Vertical Angle Theorem, angle P in triangle PBC is 111}\)

\(\text{Since the angle measures of PBC sum up to 180. We can make the following equation}\)

\(\frac{1}{2}x+\frac{1}{2}y+111=180\)

\(\text{Now we solve the equation for x + y and get }x+y=138\)

\(\text{We solved for x + y, this means we just solved for the SUM of the angle measures of ABC and ACB}\)

\(\text{Since the angle measures of triangle ABC sum up to 180, and we know that the sum of two angles is 138.}\)

\(\text{We can figure out the third angle with the following equation. }128 + n = 180\rightarrow{n}=\boxed{42}\)

i jsut checked three as well and it worked , the answer must be 3 right? since its the lowest value before they all turn into a mix of pos and neg values? Thank you so much for your help!

Yes, you stop when there are all positive values. Feel free to message me at any time!

Tertre might have typoed, yet I think he forgot the 0x^2. This is really important! Check three just to make sure. Three should work.

You're correct, six does work. 

When i solve 6 for 4 0 -25 -5 -13 i as well get all postitive vaules, dnt i need to find the lowest value which gives all postitive or all negatitive values?

Testing the values with y=1, y=2, y=3, y=4 yields negative value by using Synthetic Division. Yet, we can realize something. The -25x^3 is really messing up the positive values, so we need to find a value larger than 25. Since 6*4=24, that won't work, yet y=7 will work since 7*4=28, which is greater than 25.And yes, y=7 does work.

Thanks! That really helped!

im not sure if you remember but this is the way  i was talking about for the other question posted above, because i didnt feel like 7.25 fit properly , do u think maybe the answer would then be 13? Thank you as well for your help!

thank you for checking my answer and taking the time to help

is this one correct as well ?

I'm not sure if this is correct, but my angle relationships,

5x-7=2x+11

3x=18

x=6

Thus, angle ABP and angle EDG both meaure twenty-three(23) degrees.

The angle ABD measures 180-23=157degrees.

Thanks, tertre....!!!

Good to see you ......!!!

Not a tough as it looks

Let's suppose that  the slope of  some line  =  a/b

This theorem says that   another line will be perpendicular  if it  has  a negative reciprocal slope

The negative reciprocal of a/b   = -b/a

Farther.....if we multiply these two slope together and the result  = -1.....the lines are perpendicular

So

(a/b) (-b/a)  =

- (a/b) (b/a)  =

-  (ab) / (ba)  =

- (ab) / (ab)  =

- 1        ,,,,so....we see that they are truly perpendicular lines  !!!!!  

Posted two responses at once.

[asy]
size(220);
import TrigMacros;
rr_cartesian_axes(-3,8,-2,8,complexplane=true,usegrid = false);
pair A,B,C,D;
A = (2,1.5);
B= (4,1.8);
C = (5.3,3);
D = (3,5.3);
draw(A--B--C--D--A);
draw(A--(A + rotate(-90)*(B-A))--(B + rotate(90)*(A-B))--B);
draw(B--(B + rotate(-90)*(C-B))--(C + rotate(90)*(B-C))--C);
draw(C--(C + rotate(-90)*(D-C))--(D + rotate(90)*(C-D))--D);
draw(D--(D + rotate(-90)*(A-D))--(A + rotate(90)*(D-A))--A);
pair WW = (B + (A + rotate(-90)*(B-A)))/2;
dot(WW);
pair X = (C + (B + rotate(-90)*(C-B)))/2;
dot(X);
pair Y = (D + (C + rotate(-90)*(D-C)))/2;
dot(Y);
pair Z = (A + (D + rotate(-90)*(A-D)))/2;
dot(Z);
//draw(WW--Y,red);
//draw(X--Z,blue);
dot("$a$",A,SW);
dot("$b$",B,2*E);
dot("$c$",C,E);
dot("$d$",D,NNW);
dot("$p$",WW,E);
dot("$q$",X,S);
dot("$r$",Y,N);
dot("$s$",Z,S);
[/asy]

Divide each term by 3

1, -7/3, 5/3, -1

Remove the first term    and change any minuses to pluses

7/3, 5/3/ 1

1) take the largest value and add 1 to it  →  7/3 + 1  = 10/3

2) Sum the values =  5   .....compare this with 1 and take the larger value  = 5

Take the smaller  of  (1), (2)

The upper bound is 10/3

See this graph  : https://www.desmos.com/calculator/1mf8fqbard

However....I  think that you are using the syntethic division test to find the greatest integer that is an upper bound.....this should be 3

Test 2

2 [  3  -7   0  5  -3]

           6

  _______________

     3  -1                              we get a negative so we can stop

Test  3

3 [  3 - 7   0  5   -3]

           9   6 18   69

   _______________

     3   2    6  23  66       we ge all positives....so 3 is the largest integer upper bound 

We try the values y= 1, 2, 3, 4, 5, 6,...and so on.

For the function to have an upper bound, the coefficients of the polynomial when the original polynomial divided by (y) has to be positive. I recommend using Sythetic Division to divide the polynomial with the number.

Trying y=1, gives us a polynomial that alternates in signs. Same with y=2. Yet, y=3 gives us a polynomial with all positive coefficients. y=3: Answer