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Well...let's see

-10 (t - 1) - (t - 1)(t - 3) + 5

Let's break this up like this

-10 (t - 1) -  [(t - 1)(t - 3)] + 5

-10t  + [-10*-1] - [ t^2 - 3t - 1t + 3 ]  +   5

-10t  + [-10*-1] - [ t^2 - 4t + 3 ]  +   5

-10t  +  10  - t^2 + 4t - 3 +  5

-10t + 4t    + 10 -  3  + 5   - t^2 

-6t   +   12   -  t^2

-t^2 - 6t  + 12

You teacher made a mistake

-10t + 10 - t^2 + 3t - t - 3 + 5  should be  -10t + 10 - t^2 + 3t + t - 3 + 5

Which leads to my result  of   -10t  +  10  - t^2 + 4t - 3 +  5

Well your approach I like....I had a look and saw that you "carried" everything over to the left side in steps, and ended with:

-1 (t - 1)(t - 3) on the right side later on..

then all of that became -t^2 +4t -3..which I perfectly understand...

I was hoping to get a reasoning behind the calculation the teacher had done here;

= -10 (t - 1) - (t - 1)(t - 3) + 5 became -10t + 10 - t^2 + 3t - t - 3 + 5...

I do not understand the rule or law to change the signs like that...

I would have written:

= -10 (t - 1) - (t^2 - 3t - t + 3)..which would then become

= - 10 (t - 1) - t^2 + 3t + t - 3..which is EXACTLY what you also got..how did she get "- t" ?

I can see on the paper she used pencil to indicate she was changing the sign of the second t to a "-" inside the bracket, and the "-1" to a "+1" in this part of the sum: = -10 (t - 1) - (t - 1)(t - 3) + 5...why did she do that?...and how would this give the "-t" in any case?...I followed her calculations from thereon and everything calculates to the same end result...

The thing about this, juriemagic, is there isn't really one way to solve these things.....if we asked ten different people, we might get ten different approaches....heck....I'm not even sure mine is necessarily the best....LOL!!!!

CPill,

Thank you, I see it's slightly different solved than the teachers method...I'm going to study your method and will come back shortly...THANX!!

I see,

so it's a rule to take the second and third terms, should there be a "-" between them, together into a bracket, which obviously will have the "-" change to a "+"...and move on from there?..

Thanx for your time CPhill...

Just reverse my steps......  

There are 15 Blue Jays and 14 Orioles that wish to rest on the branches of three older trees. Each of the trees will have at least 4 Blue Jays and 2 Orioles. However, no tree may have more Orioles than Blue Jays in its branches. Determine the largest number of birds that can be in one tree.

Put   4 Jays  and 4 Orioles in each tree

So......we have   used 24  and there are 5 left  -  3 Jays and 2 Orioles

Put the 3 Jays in each of the trees   and 2 Orioles in any of  two trees

So....we will  have

5 Jays - 5 Orioles      5 Jays - 5 Orioles         5 Jays - 4 Orioles

Take   1 Jay and 1 Oriole from each of the last two trees and put them in the first tree

So we have this  arrangement

7Jays - 7 Orioles       4 Jays - 4 Orioles           4 Jays - 3 Orioles

Each tree has at least 4 Jays and  2 Orioles....and no tree contains more Orioles than Jays

And I agree with hectictar.....14 is the greatest number in any tree  !!!

Is God real?

Can't really say... I haven't met him yet.

But scientific theories and biblical information clash with each other, and both aren't sufficient enough to fully prove if God is real or not.

If you are Christian... God is real!  (At least, for you)

If you are atheist....God isn't real! (At least, not for you)

Do you know how to use the f2l method? I know the beginner's method, but youtube videos aren't very helpful when trying to learn the f2l method.

An adjective is something that describes a noun.

[ t - 2 ] / [ t - 1 ]   =  10 / [ 3-t ]   - 1   + 5 / [ t^2  - 4t + 3]

Subtract  10 / [3 - t ]  form both sides   and factor the last denominator

[ t - 2 ] / [ t - 1]   - 10  / [ 3 - t ]   =   -1  +  5 / [ (t - 1) (t - 3) ]

And we can factor a negative out of   - 10 / [ 3 - t]     so it becomes  + 10 / [ t - 3]

[ t - 2 ] / [ t - 1]   + 10  / [ t - 3 ]   =   -1  +  5 / [ (t - 1) (t - 3) ]

Subtract   5 / [ (t - 1) (t - 3) ]  from both sides

[ t - 2 ] / [ t - 1]   + 10  / [ t - 3] ]   - 5 / [ (t - 1) ( t - 3) ]   = -1

Get a common denominator on the left  = [  [t - 1 ] [ t - 3]...so we have

 ( [ ( t - 2) ( t - 3) ] + 10 [ t - 1 ]  -  5) / [ (t - 3) ( t - 1) ]    = - 1

Simplify the left side

(  t^2 - 5t + 6  + 10t - 10  - 5 )  / [ ( t - 1) (t - 3) ]  =  - 1 

( t^2 + 5t  - 9 )  /  [ ( t - 3) ( t - 1) ]  =  -1

Multiply both sides by  [ ( t - 3) ( t - 1) ] 

( t^2 + 5t  - 9 )   =   -1 [ (t - 3) ( t - 1) ]

(t^2 + 5t - 9)  = - [ t^2 - 4t +  3 ]

(t^2 + 5t - 9 ] =  -t^2 + 4t - 3      rearrange as

2t^2 + t  - 6   = 0      factor as

(2t  - 3) ( t + 2)  =  0

Set each factor to 0 and solve

2t - 3  = 0                                                 t   +   2   = 0

Add 3 to both sides                                  Subtract 2 from both sides

2t  = 3                                                        t  = -2

Divide both sides by 2

t  = 3 / 2

So    t  =  -2  , 3/2

Hi CPhil,

yes, that is the part I understand, what I do not get is how to reverse that back to the single log term?

2)

The midpoint of  (1, 2)  and  (7, 4)  \(=\,(\frac{1+7}{2},\frac{2+4}{2}) \,=\,(\frac82,\frac62)\)  =  (4, 3)

The slope between  (1, 2)  and  (7, 4)  \(=\,\frac{4-2}{7-1}\,=\,\frac26\,=\,\frac13\)

So the slope of the perpendicular bisector   =   -\(\frac31\)   =   -3

We want an equation of a line that passes through  (4, 3)  with a slope of  -3 .

In point-slope form, that is

y - 3  =  -3(x - 4)         This is the equation of the line. We need it in  y = mx + b  form.

                                   Distribute the  -3 .

y - 3  =  -3x + 12

                                   Add  3  to both sides.

y  =  -3x + 15

V =W x H x L

V =1.1 x 1.1 x 1.1

V =1.331 - 1 x 100 =33.1%=~33%. No matter what dimensions you choose for W, H, L, if you increase them ALL by 10%, the increase in volume will always be 33.1%.

1= (x/5) + 3   just subtract 3 from both sides

1 - 3 =x/5

-2 = x/5           cross multiply

x = -2 x 5

x = -10

There are 15 Blue Jays and 14 Orioles that wish to rest on the branches of three older trees. Each of the trees will have at least 4 Blue Jays and 2 Orioles. However, no tree may have more Orioles than Blue Jays in its branches. Determine the largest number of birds that can be in one tree.

This is how I would "think through" the problem.

Each tree will have at least  4  Blue Jays and  2  Orioles.

Let's go ahead and "put"  4  Blue Jays and  2  Orioles in each tree.

1^st tree:  4, 2          2^nd tree:  4, 2          3^rd tree:  4, 2          remaining:  3, 8

Now let's try to put as many of the remaining birds in the first tree as we can.

Let's put the rest of the Blue Jays in the first tree.

1^st tree:  7, 2          2^nd tree:  4, 2          3^rd tree:  4, 2          remaining:  0, 8

Now...no tree may have more Orioles than Blue Jays, so the maximum number of remaining Orioles that we can put in the first tree is  5  .

1^st tree:  7, 7          2^nd tree:  4, 2          3^rd tree:  4, 2          remaining:  0, 3

And it doesn't matter how we distribute the remaining Orioles.

The largest number of birds that can be in one tree is  14 .

Thank you, CPhill, this was really helpful. I especially liked that you told me your logic... very cool.

Jrod

1)   First we need to find the slope of  l₁ . Let's get its equation into slope-intercept form.

5x + 8y  =  -9          Subtract  5x  from both sides of the equation.

8y  =  -5x - 9           Divide through by  8 .

y  =  -\(\frac58\)x - \(\frac98\)

Now we can see that the slope of  l₁  =  -\(\frac58\)  .

Line  l₂  is perpendicular to  l₁  , so the slope of  l₂  =  +\(\frac85\)  .

Line  l₂  has a slope of  \(\frac85\)  and passes through the point  (10, 10) .

So.... in point - slope form, the equation of  l₂  is

y - 10  =  \(\frac85\)(x - 10)             We want this in the form  y = mx + b  . Distribute the  \(\frac85\) .

y - 10  =  \(\frac85\)x - \(\frac85\)(10)

y - 10  =  \(\frac85\)x - 16               Add  10  to both sides.

y  =  \(\frac85\)x - 6

Now the equation for  l₂  is in the form  y = mx + b  , where  m = \(\frac85\)  and  b = -6 .

IDK how to solve this so yeah, have fun.

6^2 / 6^-7 =6^(2 - (-7)) =6^(2 + 7) =6^9

This is equal to \({1\over u^5}*{1\over u^4}\), which simplifies to \({1\over u^9}\). 

I already answered your first question, but I'll answer this one anyway.

Fraction not orange juice = \({numberOfNotOrange\over total}={11+4\over11+4+5}={15\over20}={3\over4}\)

Fraction of Yankees cards = \({yankeesCards\over totalCards}\) = \({6\over 6+5+7}\) = \({6\over18}={1\over3}\). 

i wanna know 1x1