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Yes, CPhill's solution is also a way to get the answer!

Pretty cool!!!

Thanks CU......here's another way

Draw   PX  perpendicular to TU such that   PX intersects  TU  at W

Then  PQRTW forms an irregular pentagon

And the sum of the interior angles of any pentagon =  540

And note that the measures of angles  QRT, TWP and QPW  =  125, 90  and 90

So......the  sum of angles PQR , QRT , RTW, TWP  and QPW  = 540....so

120  + 125 + RTW + 90 +  90  = 540     simplify

425 + RTW =  540

RTW  = 540 - 425  =  115   =  RTU

thanks!

\(\text{Because angle TRS is 55, that means angle QRT is 125}\)

\(\text{We construct a line from Point R to our new point X. Let segment RX be parallel to QP and TU.}\)

\(\text{Angle PQR and angle QRX add up to 180. This means QRX must be 60}\)

\(\text{If angle QRT is 125 and angle QRX is 60, this means angle RTU is 65}\)

\(\text{Angle XRT and angle RTU sum up to 180. And since RTU is 65, this means angle RTU is }\boxed{115^{\circ}}\)

I took the PSATthe end of last year. I found it to be easy (maybe because I had already taken the ACT), and had plenty of time to double and tripple check my work on the math section, and the reading and english sections also wern't that difficult. I atually just got my scores last week.

The PSAT is supposed to be just for practice, thats strange your school is making it affect your high school classes.

I checked out a practice book from the library, and I think just understanding the format of the test and the types of questions will help you immensly. Doing a timed practice test in each of the subjects will help you get the hang of managing your time and not spending too long on difficult problems.

If you are already a good student in math and english, and you know how the test works and can manage your time, you should be able to do well on the test with out doing that much to prepare.

Nice job, CU !!!!

\(\text{We can draw a line from point B to point D to create triangle BCD}\)

\(\text{Notice how angle FBD is 90 degrees with angle FBC of 125 degrees. This means angle DBC is 35 degrees.}\)

\(\text{Since the sum of the angles of a triangle is 180, this means angle BDC is 55 degrees}\)

\(\text{Now notice how angle BDE is 90 degrees, and we have angle BDC is 55 degrees. That means angle x is }\boxed{35^{\circ}}\)

What does the C stand for, just making sure.

I plugged your equation into a calculator and it didn't work

C(14)=1.182+2.216*In(14) = {}

What are you solving for?

Okay, let x equal the overlapping area.

If the squares did not overlap, it would be a 14 x 7 rectangle, with an area of 98. But this has an area of 70.

x = 98-70 = 28

You are very welcome!

:P

Let m be the distance the squares overlap on the long edges.

First, express the area of the region the squares don't overlap in terms of the variable. (You can probabally skip this part, I'm just struggling to find more ways to use the variable :/ )

Area=2(7*(7-m))

The 2 comes from there being two rectangles, 7 is the height and 7-m is the width.

Now express the length of the total idth in terms of the variable.

14-m

With no overlap, the width would be 14, but to find the width after overlaping you have to subtract the distance they overlap

14-m=10

Set it equal to 10

-m=-4

Sutract 14 from each side

m=4

Divide by -1

Now that we have the distance they overlap, plug it into the equation from the beginning to get the answer.

2(7*(7-4))

2(7*3)

2*21

The area the squares don't overlap is 42

I like it! Very impressive!

(1998^2 - 1996^2)(1998^2 - 1995^2)   ........1998^2

_________________________________________________________

(1997^2  - 1996^2) ( 1997^2 - 1995^2)  .......1997^2

(1998 + 1996)(1998 - 1996) ( 1998 + 1995)(1998 - 1995) (1998 + 1994) (1998 - 1994).....1998^2

_____________________________________________________________________________

(1997 + 1996) (1997 - 1996) (1997 + 1995)(1997 - 1995) (1997 + 1994)(1997 - 1994).....1997^2

[ (3994)(2)]  [ (3993)(3) ] [(3992)(4) ].......[(1998)(1998)]

____________________________________________

[ (3993)(1)] [ (3992)(2)]  [ (3991)(3)]..........[(1997)(1997)]

Note that  each pair of products  in the numerator sum to 3996  and each pair of of products in the numerator sum to 3994

After cancellation of like terms we are left with

3994  *     ( 4)                 [(1998)(1998)]

                _____   .....     _____________

                ( 3991)            [ (1997)(1997)]

And....note that the sum of the non-cancelled terms in the middle fraction's numerator and denominator  =   3995

This implies.....that after complete evaluation ( and cancellations), we will be left with

3994   *    (1998)

                ______  =

                 (1997)

(2)(1997) * (1998)

_______________   =

      (1997)

2 * 1998   =

3996

As stated in my other answer, the first one can't be determined.....we can compute  R1 - (1/2)R2    but that doesn't tell us WHAT  to do with the result......

The second  is easy.......a matrix's  dimensions are expressed as  Rows X  Columns

Note that we have   4 Rows and 2 Columns

So

4 X 2

[(2/5)a + 6] * 3   = 15

(6/5)a + 18  =   15

(6/5)a  =  - 3

a  =   (-15/6)   =  (-5/2)

Check this in the other two  evaluations 

[ -1 + 6 ]  (-2)  =  [5] (-2)   =  -10

And

(-1 + 6 ] (7)  =  [5](7)  =  35

And Nirvava is correct  !!!!

A+B=260   (1)

B+C=245  (2)

C+D=270  (3)

It's system of equations

(2) multiply by -1

we get:

-B-C=-245

Add (2) to (1)

A+B=260

+

-B-C= -245 

=

A-C=15

Now let's add (3) to A-C=15

C+D=270

+

A-C=15

A+D=285 Solved 

\(a + b = 260\\ b+c = 245 \\ c+d = 270\\~\\ a+d = (a+b) +(c+d) - (b+c) = \\ 260+270-245 = 285\)

hi welcome to this math world

also correct me if i am wrong.

\(2/5a+6 \begin{bmatrix} 3&\\ -2& \\7& \end{bmatrix} = \begin{bmatrix} 15& \\ -10 & \\35 \end{bmatrix}\)

we realize that the each element in the matrix was multiplied by 5 so a must be a value that will make 2/5a+6= 5 therefore a must be a negative reciprocal of 2/5 thus we will get -5/2 because that gives us -1 and when you add -1+6=5 which satisifes the equation. 

hope this helped. :) 

\(5)If $a+b=7$ and $a^3+b^3=42$, what is the value of the sum $\dfrac{1}{a}+\dfrac{1}{b}$? \)

a  +  b  =    7     square both sides

a^2 + 2ab +b^2  = 49         (1)

a^3 + b^3  =  42         factor the left side as a sum of cubes

(a + b) (a^2  - ab + b^2)  = 42

(7) (a^2 - ab + b^2)  = 42          divide both sides by 7

a^2 - ab + b^2  =  6       (2)

Subtract  (2)  from (1)   and we have that

3ab  =  43

ab = 43/3        ( 3)

And  we can simplify    1/a  +  1/b   as     (a + b)  /  ab        

So

1/a   +  1/b   =

(a + b)  / ab    =

7  /  ( 43/3)   =

(7/1)  / (43/3)  =

(7/1) *  (3/43)  =

7 * 3  / 43   =

21 / 43

\(4)If \ $\left(a-\frac 1a\right)^2=4\ and \ a > 1,$ what is the absolute value of $a^3 - \frac1{a^3}?$ \)

Expand (a -1/a)^2  =  4   →   a^2 - 2  + 1/a^2   = 4  →  a^2 + 1/a^2  =  6   (1)

And  (a - 1/a)^2  = 4       take the positive root

a - 1/a  = 2      (2)

Factor     a^3  - 1/a^3    as  a difference of cubes

(a  -1/a)  ( a^2 + 1  +  1/a^2)

(a - 1/a) ( [a^2 + 1/a^2]  + 1)      (3) 

Sub (1) and (2)  into (3)

(2)  ( 6 + 1)

(2) (7)  = 

14

1.)

\(\text{If you want to figure out }(x-y)^2\text{, You can expand it to }x^2-2xy+y^2\)

\(\text{Notice how }xy=6\text{ now that means we can place it into our } x^2-2xy+y^2\text{ equation}\)

\(\text{So we have: }x^2-2xy+y^2\Rightarrow{x^2}-2(6)+y^2\Rightarrow{\color{green}x^2+y^2-12}\)

\(\text{Now consider one of the equations we were given: }(x+y)^2=31\)

\(\text{We can now expand the following: } (x+y)^2=31\Rightarrow{\color{red}x^2+2xy+y^2=31}\)

\(\text{Now we place }xy=6\text{ into the red equation}\)

\(x^2+2xy+y^2=31\rightarrow{x^2}+2(6)+y^2=31\rightarrow{\color{red}x^2+y^2=19}\)

\(\text{Now we combine the new red equation and the green equation together for the answer}\)

\({\color{red}x^2+y^2=19}\text{ and }{\color{green}x^2+y^2-12}\Rightarrow{19-12}=\boxed{7}\)

I do not have time to answer the other questions, but I am sure that the all involve expanding special cases and combining equations!

did it tell you why the h-value was wrong?

Im going to answer number 4 because i have no time I need to go.

Newton's third law states: For every action there is an equal and opposite force

So examples of third law

Jumping, you apply force to the ground,  ground applies an opposite force to you which makes you go higher.