JoLink

Number 5 is...  Stars?

Number 6 is Fire!

Thats All i've Figured out...

https://web2.0calc.com/questions/confusing-algebra: Link To Actuall Response

Some people seem to have some difficulty in understanding how r=7 and k=30 were obtained above!!!.

Solve for k in  equation (1) above and you should get this:

k = 6 r^2 - 35 r - 19

Solve for k in equation (2) above and you should get this:

k = 6 r^2 - 57 r + 135

Equate the above 2 equations and solve for r:

  6 r^2 - 57 r + 135 = 6 r^2 - 35 r - 19, solve for r 

 -57r +35r =-19 - 135                                                                                                  

-22r = - 154

r = 7 sub this into one of the equations above and you should get:

k = 30

[I Tried to be as neat as I could]

Let

x = the number of multiple choice question

y = the number of free response question

we know that

 3x + 8y = 55 = Equation A

x+y=15

x = 15 - y = Equation B

Substitute equation B in equation A

3[15-y] + 8y = 55

45 - 3y+ 8y = 55

5y = 55 - 45

5y = 10

y = 2

Find the value of x

x = 15 -  y

x = 15-2= 13

therefore

the answer is

the number of multiple choice question are 13

the number of free response question are 2

is it this question?

Students are given 3 minutes to complete each multiple-choice question on a test and 8 minutes for each free-response question. There are 15 questions on the test and the students have been given 55 minutes to complete it.

Because i've heard of this before.

The Answer is 2 NeverMind

Let O = (0, 0), D = (0, -8). Then B = (-12, -8) because OD = 8 and OB = 4sqrt(13), so by the Pythagorean Theorem, BD = 12

Then, since AB is a line segment passing through O, we have that A = (12, 8). Use the distance formula from (0, -8) to (12, 8) and get sqrt((12-0)^2+(8-(-8))^2) = sqrt(12^2+16^2) = 20, so AD = 20.
 

This is probably not the most efficient way, but We Still get the right answer.

The height of this trapezoid is AB and the bases are ZD and WC

We have

120 = (1/2) AB  (ZD + WC)

120 = (1/2) 12 ( ZD + 6)     multiply through by 2

240 = 12 (ZD + 6)      divide both sides by 12

20 =  ZD + 6       subtract 6 from both sides

14 = ZD

Now   angle DBW = angle BDZ

And angle ZQD = angle BQW

And since AZ = WC

Then  ZD = BW

So by AAS....triangle ZQD is congruent to triangle WQZ

And the altitude of trangle DQZ    = altitude of triangle BQW

And since twice these altitudes = AB....then each altitude = 1/2(AB) = 6

So....the area of triangle BQW   =

(1/2) BW * altitude of BQW   = 

(1/2) (14)(6) =

(1/2) (84) =

42 units^2

Sorry i read the problem wrong 

That's not possible if x+y=3 means that x+y = 5 cant happen.

You might have made a typo

-1620/9=-180

w = -1620

Your Welcome 

Δ ADB  ~ Δ AEC      

Δ ADB  ~ Δ PEB

ΔAEC  ~   Δ PEB

AE / EC  = PE / EB

AE / [CP + EP ]    = PE / EB

AE  / [ 10 + 20 ]  = 20 / 30

AE / 30  = 20 / 30  ⇒   AE   = 20

20!

Sorry i did it the long hard and painful way.

1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6

The Average is 3.24

The Answer is 3!

Thank You!

Hello There,
there are 12 combinations.

(7 x 11) + (7 + 11) = 95

7 + 11 = 18