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1. The solutions to x^2 + 5x + 3 = 0 are p and q, and the solutions to x^2 + bx + c = 0 are p^2 and q^2. Find b + c.

By Vieta

p + q  =   -5/1   =  -  5

Squaring  we  have that

p^2 + 2pq + q^2  =  25

And

pq  = 3/1   =  3      ....so....

2pq  = 6

Which implies  that

p^2  + 6  + q^2  = 25

p^2  + q^2  =  19

Which  means that in x^2 + bx +c 

-b/1   = p^2 + q^2

-b  =  19

b =   -19

And

p^2q^2  =  c /1

(pq)^2  =  c

(3)^2  = c

9 =  c

So

b + c  =   -19  + 9    =    -10

Nice, CPhill! I agree!

Thanks, brian  and CTG    !!!!

There are an infinite number of unit squares possible in a 4x4 square though, the squares do not have to be on lattice points.

Here's what I get  :

https://web2.0calc.com/questions/maths_78600

I think that is incorrect.

Since the radius of one circle is 28mm, that makes the length of the rectangle 112mm. There is no way for the area to be 117 when the length of the rectangle is 112. The rectangle isn't a really skinny one, either.

Hope this helped!

I'm just.... dying of boredom.... LOL

What is the coefficient of x^3 when 24x^4 + 6x^3 + 4x^2-7x - 5 is multiplied by 6x^3 + 3x^2 + 3x + 4 and the like terms are combined?

( 24x^4 + 6x^3 + 4x^2 - 7x  - 5)  *  ( 6x^3 + 3x^2 + 3x + 4)

The coefficient  on  x^3  will be found  as

6x^3 * 4   +  4x^2 * 3x  - 7x* 3x^2  -5*6x^3  =

[ 6* 4  + 4*3   - 7*3  - 5*6 ] x^3   =

[24  + 12  - 21   - 30  ] x^3  =

-15 x^3

So what are you doing?

In total, he has 210 cents

n + d = 30

5n + 10d = 210

multiply 5 to the first equation

then minus the first from the second one

you get 60 = 5d

12 dimes or 12 y

and 

18 nickels or 18 x

We can write two equations for this...

Where d=number of dimes and n=number of nickels.. In cents the equation will be:

10d+5n=210

d+n=30

We can use the elimination method to get

10d+5n=210

10d+10n=300

Then subtract.

-5n=-90.

So the number of nickels is 18.

That makes the number of dimes 12.

Yeah!!! My dad just watches the news and then he tells me!

f(1)   simplifies  to

3 - 7  +  2  -  b  +  1     ⇒    - 1  + b

And this   = 1

So

-1 +  b    = 1

b    =  2

I mean like what are you watching? Reading? Playing? But yeah dads are awesome!

Hint for #2: 

Factor the equation and let a and b be the roots of it. Then, compute!

If we connect the  centers  of  the circles, we will have  an equilateral triangle  with sides of  2R = 2(28) =    56mm

The  height  of this triangle, H,   can be found as

tan 60°  = H  /(1/2 side length of the triangle)

√3 = H  / 28

28√3  =  H

Then the  height  of  the rectangles  =  R  + height  of triangle  + R  =   2R + 28√3  =  2R + 28√3  = 56 + 28√3

And the  width of the  rectangle  = 4R  = 4*28  = 112

So....the area  of the  rectangle  = 

(56 + 28√3) (112)  ≈   

11703.711  mm^2  ≈  11700 mm^2

Yeah. Technically my dad. He tells me everything. LOL!

I got it but thank you though!

Thank you so much!

1)Suppose f is a quadratic function defined by f(x) = ax^2 + bx + c for some numbers a,b,and c. If g(x) = x - 2 and f(g(x)) = 2x^2 - 5x + 19 for all values of x, what is the value of a + b + c?

f(g(x)  must be    

a(x - 2)^2  + b(x - 2) +  c       expand this

a(x^2 - 4x + 4)  + bx - 2b  + c

ax^2  - 4ax  + 4a  + bx   - 2b  +  c

And we know that

ax^2  - 4ax  + 4a  + bx - 2b + c    =  2x^2  - 5x  +  19

ax^2  - ( 4a - b) x + (4a  - 2b + c)    =  2x^2  - 5x  + 19

It's obvious that  a   = 2

And   (4a - b)   = 5

So

4(2)  - b   = 5

8 - b  =  5

b   = 3

And

4a - 2b + c =  19

4(2) -2(3) + c  = 19

8 - 6  + c   =19

2 + c  =19

c  = 17

So

a + b  + c   =   2  +  3  +  17    =  22

Oh yeah forgot about playing minecraft... add that to my list. Havent seen my bff in over a week! Been Homeschooled my whole life!

School,reading,writing letters,doing math!

Thanks!!!

i got 117mmsquared as the answer its wrong what did you get

Hint:

Since the radius is 28mm, we know that the diameter is 56mm.

Therefore, the length of the rectangle will be 56mm. Knowing this, I bet you can work out the area of the rectangle :D

Hope this helped!