hectictar

The slope between  (x₁, y₁)  and  (x₂, y₂)    \(=\quad\dfrac{{\color{purple}y_2}-{\color{blue}y_1}}{{\color{purple}x_2}-{\color{blue}x_1}}\)

The slope between  (3, -2)  and  (7, 8)    \(=\quad\dfrac{{\color{purple}8}-{\color{blue}-2}}{{\color{purple}7}-{\color{blue}3}}\quad=\quad\dfrac{10}{4}\quad=\quad\dfrac52\)

The slope between  (6, 7)  and  (8, -1)    \(=\quad\dfrac{{\color{purple}-1}-{\color{blue}7}}{{\color{purple}8}-{\color{blue}6}}\quad=\quad\dfrac{-8}{2}\quad=\quad-4\)

The slope between  (1, -5)  and  (-2, 9)    \(=\quad\dfrac{{\color{purple}9}-{\color{blue}-5}}{{\color{purple}-2}-{\color{blue}1}}\quad=\quad\dfrac{14}{-3}\quad=\quad-\dfrac{14}{3}\)

The slope between  (-5, 2)  and  (-3, -7)    \(=\quad\dfrac{{\color{purple}-7}-{\color{blue}2}}{{\color{purple}-3}-{\color{blue}-5}}\quad=\quad\dfrac{-9}{2}\quad=\quad-\dfrac92\)

The slope between  (6, 3)  and  (4, 9)    \(=\quad\dfrac{{\color{purple}9}-{\color{blue}3}}{{\color{purple}4}-{\color{blue}6}}\quad=\quad\dfrac{6}{-2}\quad=\quad-3\)

Do you see how to do this? Can you do the rest?

2.

Notice that  Z  is the hypotenuse, and the hypotenuse is always the longest side.

So  Z  can't be  7 inches, because that would make it shorter than the 10 inch side.

cos( angle )  =  adjacent / hypotenuse

cos( 37° )  =  10 / Z

                                        Multiply both sides of the equation by  Z

Z cos( 37° )  =  10

                                        Divide both sides of the equation by  cos( 37° )

Z  =  10 / cos( 37° )

                                        Plug  10 / cos( 37° )  into a calculator.

Z  ≈  12.52   (inches)

3.

tan( angle )  =  opposite / adjacent

tan( X )  =  5 / 20

                                         Take the arctan of both sides of the equation.

X  =  arctan( 5 / 20 )

                                         Plug  arctan( 5 / 20 )  into a calculator.

X  ≈  14.04°

To use the sum and difference formulas, we need to find  cos(a),  cos(B),  sin(a),  sin(B),  tan(a)  and  tan(B).

Using the information given, we can find these values with the Pythagorean theorem.

\(\text{1)}\qquad\sin(a+B)\,=\,\sin( a)\cos (B)+\cos(a)\sin(B)\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,(\frac{5}{13})(-\frac12)+(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,-\frac{5}{26}-\frac{12\sqrt3}{26}\\~\\ \phantom{\text{1)}\qquad\sin(a+B)\,}=\,\frac{-5-12\sqrt3}{26}\\~\\ \text{2)}\qquad\cos(a+B)\,=\,\cos(a)\cos(B)-\sin(a)\sin(B)\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,(-\frac{12}{13})(-\frac{1}{2})-(\frac{5}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12}{26}-\frac{5\sqrt3}{26}\\~\\ \phantom{\text{2)}\qquad\cos(a+B)\,}=\,\frac{12-5\sqrt3}{26}\\~\\ \text{3)}\qquad\sin(a-B)\,=\,\sin(a)\cos(B)-\cos(a)\sin(B)\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,(\frac{5}{13})(-\frac12)-(-\frac{12}{13})(\frac{\sqrt3}{2})\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,-\frac{5}{26}+\frac{12\sqrt3}{26}\\~\\ \phantom{\text{3)}\qquad\sin(a-B)\,}=\,\frac{12\sqrt3-5}{26}\\~\\ \text{4)}\qquad\tan(a-B)\,=\,\frac{\tan(a)-\tan(B)}{1+\tan(a)\tan(B)}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}--\frac{\sqrt3}{1}}{1+(-\frac{5}{12})(-\frac{\sqrt3}{1})}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-\frac{5}{12}+\sqrt3}{1+\frac{5\sqrt3}{12}}\cdot\frac{12}{12}\\~\\ \phantom{\text{4)}\qquad\tan(a-B)\,}=\,\frac{-5+12\sqrt3}{12+5\sqrt3}\)

a² - 2ab + b²   =   (a - b)²

So....

123456789²  -  2(123456789)(123456787)  +  123456787²   =   (123456789 - 123456787)²

And....

(123456789 - 123456787)²   =   ( 2 )²   =  4

Hey, that's a good way to do it!

If   0 < x < 1 ,     \(\lfloor x \rfloor \lceil x \rceil = 0\cdot1=0\)     but     \(0\nless\sqrt0\)

If   1 < x < 2 ,     \(\lfloor x \rfloor \lceil x \rceil = 1\cdot2=2\)     and    \(1<\sqrt2<2\)

So   \(x=\sqrt2\)   is a solution to  \(\lfloor x \rfloor \lceil x \rceil = x^2\)

If   2 < x < 3 ,     \(\lfloor x \rfloor \lceil x \rceil = 2\cdot3=6\)     and    \(2<\sqrt6<3\)

So   \(x=\sqrt6\)   is a solution to  \(\lfloor x \rfloor \lceil x \rceil = x^2\)

If   3 < x < 4 ,     \(\lfloor x \rfloor \lceil x \rceil = 3\cdot4=12\)     and    \(3<\sqrt{12}<4\)

So   \(x=\sqrt{12}\)   is a solution to  \(\lfloor x \rfloor \lceil x \rceil = x^2\)

The product of the three smallest, positive, non-integer solutions  =  \(\sqrt2\cdot\sqrt6\cdot\sqrt{12}\,=\,\sqrt{144}\,=\,12\)

Honestly, they confuse me too, haha!

Here is another way to figure out the answer:

C  =  \(\frac59\)(F - 32)  =  \(\frac59\)F  -  \(\frac{160}9\)

When  F = 0 ,   C  =  \(\frac59\)(0)  -  \(\frac{160}9\)  =  - \(\frac{160}9\)

When  F = 1 ,   C  =  \(\frac59\)(1)  -  \(\frac{160}9\)  =  \(\frac59\) - \(\frac{160}9\)     Notice this value of  C  is  \(\frac59\)  more than it was when  F  was  0 .

When  F = 2 ,   C  =  \(\frac59\)(2)  -  \(\frac{160}9\)  =  \(\frac59\) + \(\frac59\) - \(\frac{160}9\)     Notice this value of  C  is  \(\frac59\)  more than the previous.

So a temperature increase of 1 degree Fahrenheit is equivalent to a temperature increase of 5/9 degree Celsius.

C  =  \(\frac59\)(F - 32)

\(\frac95\)C  =  F - 32

\(\frac95\)C + 32  =  F

When  C = 0 ,   F  =  \(\frac95\)(0) + 32  =  32

When  C = 1 ,   F  =  \(\frac95\)(1) + 32  =  \(\frac95\) + 32       Notice this value of  F  is  \(\frac95\)  more than the previous.

When  C = 2 ,   F  =  \(\frac95\)(2) + 32  =  \(\frac95\) + \(\frac95\) + 32    Notice this value of  F  is  \(\frac95\)  more than the previous.

So a temperature increase of  1  degree Celsius is equivalent to a temperature increase of 9/5 degrees Fahrenheit. (And  9/5 = 1.8). In other words, a temperature increase of 9/5 degrees (not 5/9 degrees) Fahrenheit is equivalent to a temperature increase of 1 degree Celsius.

Have you tried doodling on your paper? Here are some ideas: If you have some paper with some text, you can draw stick figures climbing or falling off of the text. If you have notebook paper with the 3 holes, you can try using those holes as circles in part of a drawing, like balloons or the center of a flower. You could draw houses. You could draw bubble or block letters. These are some things I used to do all the time.