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First one....if   ( x - 7)   is a factor, then  7  is a root....therefore....f(7)  =  0

For the second....just substitute  3 into the polynomial and evaluate.....the answer will =  p(3) = the remainder

Last one

Volume of the hemisphere  =   (2/3)pi (radius)^3

The radius  = 5 inches

So.....the volume is      (2/3)pi (5)^3  =  (250/3)pi  in^2  ≈  262  in^3

So

Volume of hemisphere / 14.44    =   262 / 14.44 ≈   18 cups

If  (x-7) is a factor, then when you calculate f(7)     '7 ' would be sustituted in for x     then (x-7 )   would be  (7-7)  and the function would = 0

Answer B

Let    y   =   x⁴  -  2x²

\(y\,=\,x^4-2x^2\\~\\ \frac{dy}{dx}\,=\,\frac{d}{dx}x^4-\frac{d}{dx}2x^2\\~\\ \frac{dy}{dx}\,=\,4x^3-4x\)

Now let's find what values of  x  make  \(\frac{dy}{dx}\)  be  0

\(0\,=\,4x^3-4x\\~\\ 0\,=\,4x(x^2-1) \)

Set each factor equal to zero and solve for  x

\(\begin{array} {lcl} 4x=0&\qquad\text{or}\qquad &x^2-1=0\\~\\ x=0&\text{or}&x^2=1\\~\\ &&x=1\quad\text{or}\quad x=-1 \end{array}\)

These are the  x  values of all the critical points.

We can look at a graph to determine which is the minimum:

For the last  one, set the listed factors to 0 and solve for x.....this  must produce 2/3

So

2x - 3  =  0     add 3 to both sides

2x  = 3            divide both sides by 2

x = 3/2     no!!!

The second and fourth answers will produce negative answers for x

But.....notice that....

3x - 2  = 0

3x = 2

x = 2/3           and "C"  is the one we want

Here's the first two

When x  = 3, y = 3.5 =  ( 3, 3.5)

When x  = 6, y = 9  =  ( 6, 9)

So....the average rate of change is like the slope between two points and is given by

[9 - 3.5 ]              5.5          55            11

_______  =       ____  =   ____   =    ___

[6 - 3 ]                  3           30              6

For the second one.....the independent variable  =  Side Length

And the dependent variable  = Area

(b)  How fast is the top of the plank moving up the wall 28 sconds after we start pushing?  

Since you push the plank for 28 seconds, then at the end of those 28 seconds the plank is no longer moving. 

Looking at it that way, the answer is zero m/sec. Do you think the question is really asking for the average rate?

6x^2 + 4y^2 = 18      divide through by 18  and we have

x^2 / 3  + 2y^2/9 =  1        and we can write this as

x^2 / 3 + y^2 / [9/2] = 1

The center is  (0, 0)

The major axis lies along the y axis

c = sqrt [ a^2 - b^2]  =  sqrt [ 9/2 - 3 ]  =  sqrt [ 3/2]

The eccentricity  =  c / a  =    sqrt [3/2] / [9/2]  =  sqrt (6) / 9

The foci  are  (0, + sqrt[3/2] )  and (0, -sqrt [3/2] )

The x intercepts are  -sqrt (3)  and (sqrt (3)

The y intercepts are  3/ sqrt (2)  and    -  3/sqrt (2)

Here's a graph :  https://www.desmos.com/calculator/nwocj2f0gf

Satisfy what??!!.

The largest possible value  is

1 + 2(2)  + 3(3)   = 

14

(p - q)^4  =  [ using the coefficients from the 4th row of Pascal's Triangle.....where the top row = row 0 )

p^4 - 4p^3q + 6 p^2q^2 - 4pq^3 + q^4

We have this possible set of moves from A to B  [  R = right   U  = up ]

( R, R, R, R, R, U,U,U,U)

Since we can place te R's in any 5 of the 9 positions  and still get from A to B....the number of total paths is

C(9,5)  = 

126

Amt after 1 year  =  1200 (1+ .03/2) + 1200 = $2418

After 18 months....she has   2418(1 + .03/2)  + 1200  = $3654.27

After two years she has   3654.27(1 + .03/2) + 1200 = $4909.08

This proof is incorrect. You are claiming that the limit that is the integral, is strictly positive because it is a limit of strictly positive numbers. That reasoning is incorrect (think about the sequence of the reciprocals of the natural numbers).

my intuition tells me that the minimum will occur when x is the average of where the zeros of each occur.

\(x = \dfrac 1 5(-12-7-3+4+8) = -2\\~\\ v = 262\)

This is the correct answer which you can confirm (and should confirm) by expanding the expression out and

recombining it into a single quadratic expression and determining the vertex of the resulting parabola.

\(x^2+7x+12 \leq 0\\ (x+3)(x+4)\leq 0\\ x \in [-4,-3]\)

\(x^2 + 5x + 6 = \left(x - \dfrac 5 2\right)^2 -\dfrac 1 4\\~\\ \text{The interval of interest is entirely to the left of the vertex of this parabola}\\ \text{and thus the value is increasing as we head towards -4}\\ \text{So the function evaluated at the endpoint, -4, will provide the maximum value}\\ \left . x^2+5x+6 \right|_{x=-4} = 16-20+6 = 2\)

The square root of a negative number produces an imaginary number.

The problem specifies it wants the real roots.

\(P=(x+2y+4z)\left(\dfrac{4}{x}+\dfrac{2}y+\dfrac{1}z\right)\\ \;\;\;\!= 12 + \dfrac{2x}{y}+\dfrac{x}{z}+\dfrac{8y}{x}+\dfrac{2y}{z}+\dfrac{16z}{x}+\dfrac{8z}{y}\\ \text{WLOG, assume } x \le y \le z\\ \text{Let }u=\dfrac{x}{y}, v = \dfrac{y}{z}, w = \dfrac{x}{z} \implies u,v,w \in (0,1]\\ \quad P\\ =12 + 2u+w+\dfrac{8}{u}+2v+\dfrac{8}{v}+\dfrac{16}{w}\\ \quad \min P\\ = 12 + 10 + 10 + 17\\ = 49\)

Remarks: The minimum is attained when x = y = z = 1.

(c) (i) Recall the definition of Riemann integration:

\(\displaystyle\int^b_a f(x) dx = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\dfrac{b-a}{n}f\left(a+\dfrac{k(b-a)}{n}\right)\)

As \(\dfrac{b-a}{n} > 0 \wedge f\left(a+\dfrac{k(b-a)}{n}\right) > 0\), every term of the Riemann sum is greater than 0.

Hence, the integral is greater than 0.

After 28 sec. the plank will be    12 - (1/4)(28)  =  12  - 7  =  5 meters from the wall

For the second part...we have that   

x^2 + y^2  =  r^2  take the derivative with respect to time

2x (dx/dt)  + 2y (dy/dt)  =  2r(dr/dt)

r = 13 m

dr/dt  = 0    [the plank length is not changing ]

dx/dt  = -1/4 m/s

At 28 sec,  x  = 5 meters     and y  = sqrt (13^2 - 5^2)  = sqrt (169 - 25)  =  sqrt (144)  = 12 meters

dy/dt  = what we are looking for

So we have

2(5m)(-1/4 m/s) + 2(12m)(dy/dt)  =  0

(10m) (-1/4 m/s)  =  - (24m)(dy/dt)

-2.5 (m^2/s) = (-24m) (dy/dt)

2.5 m^2/s  = (24m) (dy/dt)      divide both sides by 24 m

(2.5/24)(m/s) = dy/dt

(5/48)m/s  =  dy/dt

I having hard time understanding it still! Maybe if you can drow it be easier, if you can’t thats fine. Thank you 

The plank leaning against the wall forms a right triangle with one side 12 meters, and the hypotenuse 13 meters.  (We make the reasonable assumption that the wall is perpendicular to the floor or earth.)  The problem is made easier because I recognize that it's a 5 - 12 - 13 right triangle.  This saves us from having to use trig functions to determine the third side. 

The point where the plank touches against the wall is 5 meters up (That's the third side of the triangle.  If this isn't clear, draw it.). 

(a)  Moving the bottom at 1/4 meter/second for 28 seconds moves the bottom inward 7 meters, so now it's 5 meters from the wall.  

(b)  How lucky.  Since the bottom of the triangle is 5 meters from the wall and the hypotenuse is still 13 meters, we still have a 5 - 12 - 13 right triangle.  Since the top of the plank is now 12 meters up, and it started from 5 meters up, it traveled 7 meters in 28 seconds, its rate was 7/28 meters/second, which we reduce the fraction to 1/4 meter/second. 

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LOL      

If you tell someone that their answer is wrong then you need to give your reason for thinking so:

For instance,

Thanks for the time you spent helping me CPhill but I think you answer is wrong because my teacher tells me that the answer is 42.

That is just an example of what you might write.