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75  /  .60   =  125  cars  in the lot

When it hits the ground, h  =   0       ......so.....

-10t^2 + 10t + 60   = 0     multiply through  by   -1

10t^2  -10t  - 60    =  0           divide through  by 10

t^2  - t  -  6  =  0       factor

(t - 3) ( t + 2)   =   0

The  first factor set to 0  gives  us what we need

t  - 3  =   0

t  =   3  (seconds to hit the ground)

y = kx

12 = k * 8

12/ 8   =  k   =  3/2

So....when  x =  9  

y  = (3/2) (9)

y =   27 / 2   =  13.5 

Good job, swag  !!!!

HERE is a much better idea.

Go and do your own homework!

We have a rhombus

1/2  the length of  the other diagonal is    sqrt (10^2  - 5^2)   =sqrt  (100  - 25)  =sqrt (75)  = 5sqrt (3)

The diagonal lengths =  10 , 10sqrt (3)

Area   = product of  the diagonals  /  2  =    (10 * 10 sqrt (3) )   / 2  =  50 sqrt (3)

LM  =sqrt  [ 4^2  + (1 - -2)^2 ]  =  sqrt  [ 4^2 + 3^2]  =  sqrt (25) =  5

MN  =  sqrt  [ (7 - - 2)^2  ]  =  sqrt [ 9^2 ]  =  9

NQ  = sqrt   [ ( 7-3)^2  + ( 4 - 0)^2 ] =  sqrt  [ 4^2 + 4^2  ]  =    4sqrt (2)

QL =  sqrt  [ ( 3 - 1)^2  + (4 - 4)^2   ]  =  sqrt [ 2^2]   =   2

Perimeter  =  5 + 9 + 2 + 4sqrt (2)   =    16  + 4sqrt (2)   units

yˆ= 2.608x^2−115.265x+1568.065

Let s mean the can of soup and d mean the frozen dinner.

If Madison purchased 2 more frozen dinners than cans of soup, then we have the equation $d = s + 2$

If each can of soup has 500 mg , each dinner is 700 mg, and that she bought 7400 mg of sodium total, we can come up with the equation $500s + 700d = 7400$

Replacing the d in $500s + 700d = 7400$ with s + 2, we get the equation $500s+700(s+2) = 7400$

Using the distributive formula, we get

$500s+700s+1400=7400$

Simplifying this equation gives $1200s = 6000$

6000/1200 = 5

so s = 5

if d = s + 2, d must equal 7

So Madison ordered 5 cans of soup and 7 frozen dinners.

Correct me if im wrong 

(b) the line x + 3y = k - 1 intersect the curve y^2 = 2x + 5

See here  :   https://www.desmos.com/calculator/hmtcql97ih

k  =  -6

The  equation of the line is

x + 3y  =   -7

D

https://www.desmos.com/calculator/hv48lroccp

The blue line, D, has a vertex of (5,12) and it passes through (8,9)

(a) the line y = kx - 5 is tangent to the curve x^2 = 2y + 1

Using implicit  differentiation, the slope at  any point  on the  quadratic function (curve)  can be found as

2y = x^2  - 1

2y'  = 2x

y' =  x

Let  the point  where  the line is tangent to the  curve  be (x , x^2/2 - 1/2)

The line  will  intercept the point  (0,-5)

So  we  need  to  solve  the slope equation

( x^2/2  - 1/2  -  - 5)  / ( x - 0)  = x

(x^2/2 - 1/2 + 5)  /  x   =   x        multiply through  by x

x^2/2 + 9/2  = x^2                 multiply through by 2

x^2 + 9  = 2x^2        rearrange  as

x^2  - 9   =  0

x^2  = 9             take both roots

x = 3      or   x  = -3       =     the two values for k

The  lines  are

y = 3x  - 5      and    y   = -3x  -5

See here :   https://www.desmos.com/calculator/npleonadqb

if you add two polynomials together, the degree of that new polynomial will be the greater degree between the two polynomials

so the answer is 11, the degree of p(x)

This is too many unrelated questions for ONE post..... Please post unrelated questions seperately ,,,,    

This is a graph of an odd  function

An odd  function  has  the property  that   f(-a)  + f(a)   = 0

Therefore

u(-2.72)  + u(2.72)  + u(-0.81)  + u(0.81)   =   

             0                +                 0       =   

                                0

h = kl  

h = 25    l = 50

   25 = k(50)     means the proprtionality constant, k, is   1/2

so    h = 1/2 L

        h = 1/2 (10) = 5     when L = 10

A pipe with inside diameter 10'' is to carry water from a reservoir to a small town in an arid land. Neglecting the friction and turbulence of the water against the inside of the pipes, what is the minimum number of 2''-inside-diameter pipes of the same length needed to carry the same volume of water to the arid town?   

The volume is a product of the cross-section area. 

A pipe with 5" radius has an area of 25π.  

A pipe with 1" radius has an area of   1π.  

So it looks like it'd take 25 of those 2-inch pipes to equal the volume carried by the 10-inch pipe.  Amazing.  

_.

Add all of these together :

    TWO triangular ends     area =  1/2 b h   =  1/2  6 x 4

     TWO rectangular sides    area   5 x 9

      ONE bottom                        area = 6 x 9           ( the bottom may or may not count as surface area in this problem......!)

Let's see  if we have a pattern

Term   

1  = 13 

2  =  1^3  + 3^3   =  28

3  =   2^3  + 8^3   =  520

4  =  5^3 +  2^3   =  133

5  = 1^3 + 3^3 + 3^3  = 55

6  = 5^3 + 5^3  =  250

7  = 2^3 + 5^3  =  133

8 = 1^3 + 3^3 + 3^3  =  55

9 = 5^3 + 5^3  =  250

Notice that.....after  the 3rd term, the pattern repeats  in cycles  of  3

The 100th term  is   100 mod 3 = 1  =   1st term  in the pattern =  133

Y = f(x) + 4  just shifts the graph of the function up 4 untis

        f(3)  would then be  4 + 4 = 8

https://web2.0calc.com/questions/a-few-questions-i-need-help-on#r1