hectictar

Since  (a, b)  is on the graph of  y = f(x),  we can say that:   f(a) = b

Let's plug in (a-6)/3  for  x  into the equation for the second graph.

y  =  -3 * f( 3 [ (a-6)/3 ] + 6)  -2   =   -3 * f(a) - 2   =   -3b - 2

When  x = (a-6)/3 ,  y = -3b - 2  so we can say that the point  $(\frac{a-6}{3}, -3b-2)$  must be on the second graph.

$g(x)=(x^2-x+1)e^x$

The function is continuous in the interval [-4, 1] so, to find its maximum value, we only have to examine the endpoints of the interval and the points at which the slope (derivative) is 0.

First let's find all the values of  x  that make the derivative equal  0 .

$g'(x)=(x^2-x+1)(e^x)+(e^x)(2x-1)$

Set the derivative equal to 0 and solve for  x :

$(x^2-x+1)(e^x)+(e^x)(2x-1)=0\\~\\ e^x(\ (x^2-x+1)+(2x-1)\ )=0\\~\\ e^x(\ x^2+x\ )=0\\~\\ e^x(x)(x+1)=0$

Set each factor equal to zero and solve each equation for  x:

$\begin{array}{} e^x=0&\qquad\text{or}\qquad &x=0&\qquad\text{or}\qquad& x+1=0\\~\\ \text{no solution}&&x=0&&x=-1 \end{array}$

So only the following x values have the possibility to produce the maximum  g  in the interval:  -4, -1, 0 1

$\begin{array}{} g(-4)&=& ((-4)^2-(-4)+1)e^{-4}&=& \frac{21}{e^4}& \approx&0.385\\~\\ g(-1)& =& ((-1)^2-(-1)+1)e^{-1}& =&\frac{3}{e^1}&\approx& 1.104\\~\\ g(0)&=& ((0)^2-(0)+1)e^{0}& =& (1)1& =& 1\\~\\ g(1)&=& ((1)^2-(1)+1)e^{1}& =& (1)e^1& \approx& 2.718 \end{array}$

The maximum value of  g  in the interval  [-4, 1]  is  2.718  and it occurs when  x = 1

$3\lt\sqrt{2x}\lt4\\~\\ 3^2\lt(\sqrt{2x}\ )^2\lt4^2\\~\\ 9\lt2x\lt16\\~\\ \frac92\lt\frac{2x}{2}\lt\frac{16}{2}\\~\\ 4.5 \lt x \lt 8$

The integers between 4.5 and 8 exclusively are:

5, 6, 7

So there are 3 different integer values of  x  that satisfy the condition.

$\frac{Bx-11}{x^2-7x+10}\ =\ \frac{A}{x-2}+\frac{8}{x-5}\\~\\ \frac{Bx-11}{(x-2)(x-5)}\ =\ \frac{A}{x-2}+\frac{8}{x-5}\\~\\ \frac{Bx-11}{(x-2)(x-5)}\cdot(x-2)(x-5)\ =\ \Big(\frac{A}{x-2}+\frac{8}{x-5}\Big)\cdot(x-2)(x-5)\\~\\ Bx-11=\frac{A(x-2)(x-5)}{x-2}+\frac{8(x-2)(x-5)}{x-5}\\~\\ Bx-11\ =\ A(x-5)\ +\ 8(x-2)$

Assuming the previous equation is true for all values of  x ,  it must be true when  x = 0 , so we can say:

$B(0)-11\ =\ A(0-5)+8(0-2)\\~\\ -11\ =\ A(-5)-16\\~\\ 5\ =\ A(-5)\\~\\ A\ =\ -1$

And it must also be true when  x = 5  so we can say:

$B(5)-11\ =\ A(5-5) + 8(5 - 2)\\~\\ 5B-11\ =\ 24\\~\\ 5B\ =\ 35\\~\\ B\ =\ 7$

So   A + B   =   -1 + 7   =   6

A quadratic function with zeros at  a  and  b  is:

y  =  (x - a)(x - b)

Then we can multiply out the right side of that equation, or "FOIL" it, to get:

y  =  x² - bx - ax + ab

And then continue to simplify the right side until it looks like a nice quadratic equation.

Using this "template", we can find a solution to each of these problems.

a)    y   =   (x - 3)(x - 5)   =   x² - 5x - 3x + 15   =   x² - 8x + 15

b)    y   =   (x + 4)(x - 3)   =   x² - 3x + 4x - 12   =   x² + x - 12

c)    y   =   $(x-\frac12)(x-\frac23)\ =\ x^2-\frac23x-\frac12x+\frac13\ =\ x^2-\frac76x+\frac13$

d)    In this case, the one root is  2 + √3  and the other is  2 - √3 , so the quadratic equation is:

        y  =  ( x - (2 + √3) )( x - (2 - √3) )

        y  =  ( x - 2 - √3 )( x - 2 + √3 )

        y  =  x² - 4x + 1

Can you figure out the last one? It might be the trickiest one, so if you need more help on it just ask!

The sum of the measures of the exterior angles of a polygon is 360°

Since this polygon is regular, each exterior angle measures 18°

Let  n  =  the number of exterior angles  =  the number of sides of the polygon

So we can say:

18º * n  =  360º

n  =  360º / 18º

n  =  20

$\phantom{=}\quad\dfrac{x^1\cdot x^2\cdot x^3\cdots x^{10}}{x^2\cdot x^4 \cdot x^6 \cdots x^{20}}\\~\\ {=}\quad\dfrac{x^{1+2+3+\dots+10}}{x^{2+4+6+\dots+20}}\\~\\ {=}\quad\dfrac{x^{55}}{x^{110}}\\~\\ {=}\quad x^{55-110}\\~\\ {=}\quad x^{-55}\\~\\ {=}\quad \dfrac{1}{x^{55}}$

If  x = 2  we get:

$\dfrac{1}{2^{55}}\ =\ \dfrac{1}{36028797018963968}$

Let:

f₁(x)  =  ax + 3

f₂(x)  =  x + 5

f₃(x)  =  2x - b

If the function is continuous, that means  f₁(2)  must equal  f₂(2)  and also that  f₃(-2)  must equal  f₂(-2)

f₁(2)  =  f₂(2)

a(2) + 3  =  2 + 5

2a + 3  =  7

2a  =  4

a  =  2

f₃(-2)  =  f₂(-2)

2(-2) - b  =  -2 + 5

-4 - b  =  3

-b  =  7

b  =  -7

And so

a + b  =  2 + -7  =  -5

Oh thank you!! What a cool surprise

And thank you Melody! That is a super cute picture ☺️☺️☺️

Let  c  be the number of chocolate puffs he baked and let  s  be the number of strawberry puffs he baked.

Since he baked a total of 1060 puffs, we can make the following equation

c + s  =  1060

Let  x  be the number of puffs he gave away for each type.

So since he was left with 2/7 of the chocolate puffs after giving away  x  chocolate puffs, we can say:

c - x  =  $\frac27$c

And since he was left with 1/5 of the strawberry puffs after giving away  x  strawberry puffs, we can say:

s - x  =  $\frac15$s

The total number of puffs left is going to be 1060 - 2x. So now we need to find  x  by solving this system of three equations.

s - x  =  $\frac15$s          ⇒          s - $\frac15$s  =  x          ⇒          $\frac45$s  =  x          ⇒          s  =  $\frac54$x

c - x  =  $\frac27$c          ⇒          c - $\frac27$c  =  x          ⇒          $\frac57$c  =  x          ⇒          c  =  $\frac75$x

c + s  =  1060              Now we can substitute  $\frac54$x  in for  s  and  $\frac75$x  in for  c

$\frac75$x  +  $\frac54$x  =  1060

$\frac{53}{20}$x   =   1060

x   =   1060 * $\frac{20}{53}$

x   =   400

And so the total number of puffs left  =  1060 - 2x  =  1060 - 2(400)  =  1060 - 800  =  260