CPhill

You have two different hourly rates .......????

We can do this by working out the areas of the rectangles  = base * height

Rectangle   6 -10 .....base  = 4  and height  = 3....area  =  12

Rectangle 10-12  ....base  = 2  and height  = 8....area  =  16

Rectangle 12 - 14  ....base  = 2    and height  = 11.....area  = 22

Rectangle 14 -15......base  = 1 and height = 23...area  = 23

Rectangle 15 - 16 .....base  = 1  and height  = 19  ....area  = 19

Rectangle 16 - 20...base  = 4  and height  = 2....area  = 8

Number  of mice  below 10g  =

[ Area  of rectangle 6-10] / [ Area of all rectangles]  * 1000  =

[ 8 ] / [ 12 + 16 + 22 + 23 + 19 + 8 ]  * 1000  =

[ 8 ]  / [100] * 1000   =     8000 / 100  =    80  mice

Add the numbers  and divide this sum by 7.....that will be the average per day...

Second one :

Put these  in the form   ax^2  + bx + c  = 0

The discriminant  is given by   [ b^2 - 4ac ]...if this is > 0, we will have two real roots

3x^2 - 4x + 7  =  0

discriminant  =  (-4)^2 - 4(3)(7)  =  16 - 84  ....not > 0....so...no two real roots

1x^2 - 2x + 8  = 0

discriminant  =  (-2)^2 - 4(1)(8)  =  4 - 32  ....not > 0...so....no two real roots

5x^2 + 6  =  0

discrimant  =  (0)^2  - 4(5)(6)  = - 120  ....not > 0....no real roots

4x^2  - 9  = 0

discriminant  =  (0)^2 - 4(4)(-9)  = 36 > 0  ....  so....two real roots

2x^2 + 3x  - 4  = 0

discriminant = (3)^2  - 4(2)(-4)  =  9 + 32  > 0...so...two real roots

1x^2 - 5x - 3  = 0

discriminant =  (-5)^2 - 4(1)(-3)  = 25 + 12  > 0....so.....two real roots

DC  = BC  ...  thus... they will  both be  =  DB  / √2    [ the diagonal of a square is always √2 * side length ....so...the side length  =  diagonal / √2  ]

So

If  DB  = 4  then   DC = BC  = 4 / √2  = [ rationalize the denominator] = 4√2 / 2 = 2√2...so...false

If DB  = 14  then DC = BC  =  14 / √2   =  [rationalize the denominator]  =  14 * √2  / [ √2 * √2 ] =

14√2 / 2   =  7√2....so......this is true

If DB  = 4√2  then DC = BC  =  4√2 / √2  =  4....so....true

If DB  = √10  then DC = BC  = √10 /√2   =  √5   ....so...true

If DB = 15 then  DC = BC  =  15 / √2  =  15√2 / 2.....so true

If DB  = √5 then  DC = BC  = √5 / √2   =   √5 * √2  / 2  =  √10 / 2...so....false

Domain...correct  !!

Range  [ we are asking...what y values are covered by the graph???]....note  that the graph goes from the "low" point of -inf  up to the "high" point  of .056....so  (-inf, .056)

Rel max(s)  =  ( -1.107, .056) U ( .838, -1.927)

Rel min(s)  =  (.27, - 2.13)

End behavior  =   -inf at both ends

Inc intervals  =  ( -inf, -1.107)  U (.27, .838)

Dec Intervals =  ( -1.107, .27)  U (.838, inf)

Here's the graph : https://www.desmos.com/calculator/s7t9d3yfdb

Domain  ( -inf, inf)

Range [ ≈ -5.848, inf )

Rel Max  =  ( 1.366, -.652)

Rel Mins  =  ( -.366, -5.848)  and  ( 2, -1)

End behavior   =  +inf  at both ends

Increasing intervals  (-.366, 1.366)  U (2, inf)

Decreasing Intervals (-inf, -.366) U (1.366, 2)

Here's the graph, RP  : https://www.desmos.com/calculator/t29xpgopwz

Yeah...I figured so RP......

She's trying as hard as she can to understand this stuff....maybe she is home-schooled...if so....I can see why it's difficult  !!!

Don't judge someone until you have walked in their shoes......

Domain, Range :   (-inf, inf)

Relative max :  (0,6)

Relative min : (2.333, - 6.704)

End behavior :  - inf  at left end.....+ inf  at right end

Increasing intervals  = (-inf, 0)  U (2.333, inf)

Decreasing interval = (0, 2.333)

Here's the graph, RP :  https://www.desmos.com/calculator/ovzplqomhc