hectictar

Let the length of the hypotenuse in inches  =  h

Then by the Pythagorean theorem,

5²  +  4.5²   =   h²

25 + 20.25   =   h²

45.25   =   h²

√[ 45.25 ]   =   h

The perimeter of a triangle is the sum of the length of all its sides. So...

perimeter in inches   =   5  +  4.5   +  √[ 45.25 ]

perimeter in inches   =   9.5  + √[ 45.25 ]

perimeter in inches   ≈   16.227

GCF( LCM(28, 14), LCM(7, 14) )

=  GCF( 28, 14 )

=  14

An equation for the parabola with its vertex at  (5 -3)  and a vertial axis of symmetry is:

y  =  A(x - 5)² - 3

Since  (2, 8)  is a solution/point on the graph, we can plug in  2  for  x  and  8  for  y  to find  A:

8  =  A(2 - 5)² - 3

11  =  A(2 - 5)²

11  =  9A

11/9  =  A

So we know that the graph of the following equation is the parabola we want:

\(y=\frac{11}{9}(x-5)^2-3\)

Now we just need to get it into the form of   y = ax² + bx + c

\(y=\frac{11}{9}(x-5)(x-5)-3\\~\\ y=\frac{11}{9}(x^2-10x+25)-3\\~\\ y=\frac{11}{9}x^2-\frac{110}{9}x+\frac{275}{9}-3\\~\\ y=\frac{11}{9}x^2-\frac{110}{9}x+\frac{275}{9}-\frac{27}{9}\\~\\ y=\frac{11}{9}x^2-\frac{110}{9}x+\frac{248}{9}\)

Now it is in the form we want, and so we can see that

a + b + c   =   \(\frac{11}{9}+-\frac{110}{9}+\frac{248}{9}\ =\ \frac{11-110+248}{9}\ =\ \frac{149}{9}\)

Is this your integral?

\(\displaystyle\int9(3x+10)^6 dx\)

If so....

Let's pick   u = 3x + 10   which means  du = 3 dx   and then substitute these in like this:

\(\phantom{=\quad}\displaystyle\int9(3x+10)^6 dx\\~\\ {=\quad}\displaystyle\int3\cdot3(3x+10)^6 dx\\~\\ {=\quad}3\displaystyle\int(3x+10)^6 \cdot 3dx\\~\\ {=\quad}3\displaystyle\int u^6 \cdot du \\~\\ {=\quad}3\displaystyle\int u^6 du\)

                                           Now we can use the rule which says  \(\int u^n du\ =\ \frac{u^{n+1}}{n+1}+c\)    to get...

\({=\quad}3\cdot\frac{u^7}{7}+c\\~\\ {=\quad}\frac37u^7+c\)

                                           Now let's substitute  3x + 10  back in for  u

\({=\quad}\frac37(3x+10)^7+c\)

Also, one way to get a nice picture of the graph is to click the share button and then click "Export Image" 

If you feel comfortable using Desmos, I just made this template to create Venn Diagrams with:

I used "paintbrush" which is basically an equivalent to MS Paint for Mac

(I'm not sure if MS Paint has a feature to adjust the opacity of a color...it has been a while since I have used that!)

There are 5 dogs that haven't learned any tricks.

(If you would like an explanation on how to make this Venn diagram please ask )

The question is how to show that the above two products are equal to each other.

I don't know if this is the best way, but this is how I would do it:

\(\large {\prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1+m}^{k=n+m}a_k\)     because  \( \prod\limits_{k=1+m}^{n+m}a_k\)   means the same as  \( \prod\limits_{k=1+m}^{k=n+m}a_k\)

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k-m=1}^{k-m=n}a_k\)      by subtracting  m  from both sides of each equation

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k-m=1}^{k-m=n}a_{(k-m+m)}\)         since  k - m + m  =  k

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{{\color{RubineRed}k-m}=1}^{{\color{RubineRed}k-m}=n}a_{({\color{RubineRed}k-m}+m)}\)

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1}^{k=n}a_{(k+m)}\)       because we can replace the pink text with whatever we want

\(\large \phantom{ \prod\limits_{k=1+m}^{n+m}a_k}\ =\ \prod\limits_{k=1}^{n}a_{(k+m)}\)

_