hectictar

(a)

Let the length of side AB  =  c

and let  m∠ABC  =  B

and let  m∠BAC  =  A

We're given that   \(CM = \frac{1}{2} AB\)   so

CM  =  c / 2

And because  M  is the midpoint of side AB,

AM  =  c / 2

BM  =  c / 2

Therefore both  △AMC  and  △BMC  are isosceles triangles. Because base angles are congruent,

m∠MCA  =  A

m∠MCB  =  B

And now we can see that

m∠ACB  =  A + B

The sum of the interior angles of △ABC  =  180°

(A)  +  (A + B)  +  (B)  =  180°

2A + 2B  =  180°

2(A + B)  =  180°

A + B  =  90°

m∠ACB  =  90°

Find t if the expansion of the product of x^3 - 4x^2 + 2x - 5 and x^2 + tx - 7 has no x^2 term.

If we add up all the tems that will have  x²  in them, we would get  0

(-4x²)(-7) + (2x)(tx) + (-5)(x²)  =  0

28x² + 2tx² - 5x²  =  0

                                       Factor  x²  out of all the terms on the left side.

x²( 28 + 2t - 5 )  =  0

                                       We only want to know what  t  will make this equal  0

28 + 2t - 5  =  0

23 + 2t  =  0

2t  =  -23

t  =  -11.5

Suppose f is a polynomial such that f(0)=47, f(1)=32, f(2)=-13, and f(3)=16. What is the sum of the coefficients of f?

Depending on your definition of "coefficient", the answer is either

f(1)  =  32

which does count the constant term

or

f(1) - f(0)  =  32 - 47  =  -15

which does not count the constant term

Using the tactic that you had been using.....

-\(\frac12\)x + 3  =  -x + 7

                                            Multiply both sides of the equation by  -2

-2( -\(\frac12\)x + 3 )  =  -2( -x + 7 )

                                            Distribute the  -2  to each term

x - 6  =  2x - 14

                                            Add  14  to both sides of the equation.

x - 6 + 14  =  2x - 14 + 14

x + 8  =  2x

                               Subtract  x  from both sides.

x + 8 - x  =  2x - x

                               On the left,  we have  x - x = 0

8  =  2x - x

                               On the right,  we have  2x - 1x = 1x  because  2 things - 1 thing = 1 thing

8  =  1x

8  =  x

=   1 + 3 + 5 + . . . + 2017 + 2019 - 2 - 4 - 6 - . . . - 2016 - 2018

=  (1 - 2) + (3 - 4) + (5 - 6) + . . . + (2015 - 2016) + (2017 - 2018) + 2019

=  (1 - 2) + (3 - 4) + (5 - 6) + . . . + (2015 - 2016) + (2017 - 2018) + 2019

=  -1   +   -1   +   -1   + . . . +   -1   +   -1   +   2019

Now the question is how many  -1 's  are there?

the number of  -1 's   =   2018 / 2   =   1009

So the sum in question   =   1009( -1 )  +  2019   =   1010

sin( angle )  =  opposite / hypotenuse

sin( R )  =  TS / 5

2 / 5  =  TS / 5

2  =  TS

Now we can find the length of RS using the Pythagorean Theorem.

2² + RS²  =  5²

RS²  =  5² - 2²

RS²  =  21

RS  =  √[ 21 ]

sin( angle )  =  opposite / hypotenuse

sin( T )  =  RS / 5

sin( T )  =  √[ 21 ] / 5

sin A  =  2 cos A

                               Divide both sides of the equation by  cos A

\(\frac{\sin A}{\cos A}\)  =  2

                     And  \(\frac{\sin A}{\cos A}=\tan A\)  so we can substitute  tan A  in for  \(\frac{\sin A}{\cos A}\)

tan A  =  2

1)   P(x)  =  P(0) + P(1)x + P(2)x²

Plug in  -1  for  x  to get:

Plug in  2  for  x  to get:

So we have found that     P(x)  =  -1 - x + x²

\(S\ =\ \frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{1}{\sqrt{n}+\sqrt{n+1}}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{1}{\sqrt{n}+\sqrt{n+1}}\cdot \frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{\sqrt{n}-\sqrt{n+1}}{(n)-(n+1)}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{\sqrt{n}-\sqrt{n+1}}{-1}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\sqrt{n+1}-\sqrt{n}\\~\\ S\ =\ {\color{}\sqrt2}-\sqrt1\ {\color{}+\sqrt3-\sqrt2+\sqrt4-\sqrt3+\sqrt5-\sqrt4+\dots+\sqrt{99}-\sqrt{98}}+\sqrt{100}\ {\color{}-\ \sqrt{99}}\\~\\ S\ =\ {\color{gray}\sqrt2}-\sqrt1\ {\color{gray}+\sqrt3-\sqrt2+\sqrt4-\sqrt3+\sqrt5-\sqrt4+\dots+\sqrt{99}-\sqrt{98}}+\sqrt{100}\ {\color{gray}-\ \sqrt{99}}\\~\\ S\ =\ -\sqrt1+\sqrt{100}\\~\\ S\ =\ -1+10\\~\\ S\ =\ 9 \)_

(a)

\(\sqrt{x^2+y^2}\ =\ 5\)

The solutions to this equation are all points with a distance of  5  from the origin.

So this is the equation of a circle with a radius of 5 centered at the origin.