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See diagram below:   You need 3 m of depth ( 2.5 m  +  .5 meters ) 

       Yellow line is tide at 2.5 m      Blue line is 5:30 PM      (Intersection is their situation presently)

       Green line is depth 3.0 m       Black line is 10 PM (when the tide causes depth to next be at 3.0 m)

I know what you mean ...

When I set the ambiguity filter to 0.5, the question changed to: How do you thin out the population of an overcrowded school?   

Solution: Put a road through the school. The slowest students and those who do not pay attention will be the first to go. 

(It’s easy to misapply Occam’s Razor)

GA

--. .-

Find f(y) + g(y).

Hello Guest!

\(f(y)+g(y)=y^4 - 3y^3 + y - 3+ y^3 + 7y^2 - 2\)

\(f(y)+g(y)=y^4-2y^3+7y^2+y-5\)

  !

What is b+c?

Hello Guest!

\((z-(-7 + 2i)) * (z-(7 - 2i))=0\\ (z-(7-2i))(z+(7-2i))=0\\ z^2-(45-28i)=0\)

\(z^2 + bz + c = 0\)

\(b=0\\ c=-45+28i\\ \color{blue}b+c=-45+28i\)

  !

Actually, you are right.   I don't know that.  I can see it is close to that but it may not be exact ....

Thanks for picking that up.   I 'll think about it a bit more ...

Is the point (1, 3) a solution to the linear equation 5x-9y=32 ?

Hello Guest!

No, but the point (1, -3) is a solution to the linear equation 5x-9y=32.

  !

Area = 1/2(12 * 18 - 5 * 12) = 78

Hi there!  The area is 62 square inches.  Hope this helps!

3a  + 18 ==39,      

[3a  +  18] / 12==3.25, solve for a

There are: 9 twos  and 3 sevens

By the law of exponents, if \(3^a=2, \) and \(3^b=5,\) then \(3^{a+2b}=2 \cdot 5 \cdot 5 = 50.\) Now we know that \(50 \cdot 3 = 150, \) so \(3^{a+2b+1}=150,\) so in terms of \(a\) and \(b, \) \(\boxed{x=a+2b+1}\)

Hello Guest,

Here is my attempt:

There are 16 people who own cats, and 5 people who own cats and dogs, so there are 11 people who only own cats. Now there are 9 people who own dogs or both, so the answer would be \(11 + 9 = \boxed{20}\) people own a cat or a dog

Let x cookies into Bottle B first.

Now, Bottle A cookies = 3/4x

24 from Bottle A to Bottle B

∴ No A cookies in Bottle A = 3/4x-24

                                     B = x + 24

On the given condition

          1/4 (x+24)=3/4x-24

       => x+24=3x-96

      => 2x=96+24=120

      => x=120/2=60

      => x=60

      There were 60 cookies in Bottle B

                ∴ in Bottle A = 60 * 3/4 = 45 cookies

     Now (60-45) = 15

Ans, There were 15 cookies more in Bottle B than Bottle A out first.

Let the number of cookies in each tins is x.

Then the number of cookies in 8 tins = 8x.

Given, he took 15 cookies from each tin therefore he took

8 * 15 = 120 cookies from 8 tind

Hence, the total number of cookies left in 8 tins = 8x - 120

Given the left cookies in 8 tins is equal to the total number

of cookies in 2 of the tins at first

That is left cookies in 8 tins = 2x.

Hence, 8x - 120 = 2x => 6x = 120

                                  => x = 20

Hence, 20 cookies are there in each tin at first.

First of all, thank you! 

But how do you know the foci must be at (-1,0) and (1,0)? Why does this minimize the area... 

w = (2/5) s
(3/5) s = others = bike + bus
bus/bike = 7/2
bike = (w - 224)
------------------------
bus = 7 (w-224) / 2 = 7[(2/10)s - 112 ] 
bike = (2/5)s - 224
sum = (3/5) s
so
(3/5) s = (2/5)s - 224 + 7[(2/10)s - 112 ]
6 s = 4 s - 2240 + 14 s - 784
-12 s = -3024
s = 252

Let us consider that there were x blue pens and y green pens in the box.

Given that if 27 green pens were taken out of the box, there would be 1/3 as many blue pens as green pens. Then we have

x = (y-27)/3

This gives 3x - y = -27..............(1)

Also if 40 blue pens were added into the box, there would be 1/2 as many blue as green pens. Then we have (x+40) = y/2.

This gives 2x - y = -80................(2)

Now subtracting (2) from (1), 

we have x = 53 

From (1) we have y = 186.

Therefore, there was 53 blue pens and 186 green pens in the box.

Let's include a slightly more helpful version of your graph here.

I haven't done it yet and I always have to look up the 'rules' when I am dealing with ellipses

but it is pretty clear that the focii are (-1,0) and (1,0)

I'd take it from there.  I might do it later.

idk Ginger.

It all looks good but I m not sure you used the right ambiguity setting.

If you miscalculated that then your whole answer is in dire jeopardy    

The sum of all possible values of k is 38.

The probability is 5/28.

Solution:

After using a babble translator with an ambiguity filter setting of 0.6

and applying novacula Occami: lex parsimoniae (Ocham's razor: law of parsimony)

I find that ...

...This is solvable by using the Secant-Tangent Product Theorem

(Total secant segment length)*(external secant segment length) = (length of tangent segment)²

Bridge portion of the road (circle portion secant length) = 5 km

Road from the bridge to where the new road will be (external portion of secant length) = 4 km

Sum of above lengths (extended secant length) = 9 km

External portion times extended length = (4*9) =36 (Square of Tangent length)

Tangent length = sqrt(36)=  6 km (How long the road will be constructed)

GA

--. .-

can you explain how you got to this answer?

2/5x + 28 = 3/4x - 28

x = 160

Josh had $120             

3/4(160) = 120     

  120 - 28 = 92     

  2/5(120)=48

Ken had  $64             

  2/5(160) = 64         

 64 + 28 = 92       

Josh money =3/4(64) = 48

the polynomial is z^4 - 2z^3 - 11z^2 + 18z - 9 = 0.

The roots are 0, i, and -i.