hectictar

27.

K  =  area of triangle  =  \(\frac12\) ( base )( height )

Let the base  =  a

and the height  =  c sin B

Let's get  c  in terms of what we know using the Law of Sines

\(\frac{c}{\sin C}\,=\,\frac{a}{\sin A}\\~\\ c\,=\,\frac{a\sin C}{\sin A}\)

So the height  =  c sin B  =  (\(\frac{a\sin C}{\sin A}\)) sin B  =  \(\frac{a\sin B\sin C}{\sin A}\)

K  =  area of triangle  =  \(\frac12\) ( base )( height )  =  \(\frac12\) ( a )( \(\frac{a\sin B\sin C}{\sin A}\) )   =   \(\frac{a^2\sin B\sin C}{2\sin A}\)

\(K\,=\,\frac{a^2\sin B\sin C}{2\sin A}\)

We can simplify     x²  + x²     if that is what you're asking....

Just like  1 apple  +  1 apple  =  2 apples,     x²  +  x²  =  2x²

Second one:

point  P  =  (-7, 3)

point  Q  =  (2, 1)

It looks like you have found the same coordinates out to the side?

Let  a,  b,  c,  and  d  be real numbers such that  a > b  and  c > d .

\(\text{__________________________________________________________________________________}\)

(A)     a + c  >  b + d

We know.....

a  >  b_____and_____c  >  d

Add  d  to both sides of the first inequality.

a + d  >  b + d

If we substitute  c  for  d  on the left side it will make the left side bigger,

and the inequality will still be true. So it is always true that

a + c  >  b + d

\(\text{__________________________________________________________________________________}\)

(B)     2a + 3c  >  2b + 3d

By the same logic from part (A), we can conclude that it is also true that     2a + 3c  >  2b + 3d

\(\text{__________________________________________________________________________________}\)

(C)     a - c  >  b - d

Counterexample:     Let   a = 2 ,  b = 1 ,  c = 2 ,  d = 1

a  >  b    and    c  >  d    but it is not true that    a - c  >  b - d    because    2 - 2  >  1 - 1    is not true.

\(\text{__________________________________________________________________________________}\)

(D)     ac  >  bd

Counterexample:     Let   a = 5 ,  b = 1 ,  c = -1 ,  d = -2

a  >  b    and    c  >  d    but it is not true that    ac  >  bd    because    (5)(-1)  >  (1)(-2)    is not true.

\(\text{__________________________________________________________________________________}\)

(E)     a² + c²  >  b² + d²

Counterexample:     Let   a = 2 ,  b = -5 ,  c = 2 ,  d = 1

a  >  b    and    c  >  d    but it is not true that    a² + c²  >  b² + d²    because    2² + 2²  >  (-5)² + 1²    is not true.

\(\text{__________________________________________________________________________________}\)

(F)     a³ + c³  >  b³ + d³

If     a  >  b     then     | a³ |  >  | b³ |     and     a³  >  b³

If     c  >  d     then     | c³ |  >  | d³ |     and     c³  >  d³

By the same logic from part  (A) , we can conclude that it is also true that     a³ + c³  >  b³ + d³

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:

Thanks! I've learned a thing or two from your answers too!

Let the number of girls  =  g

Let the number of boys  =  b

b + g  =  20

b  =  20 - g

probability that a girl is chosen  =  \(\frac23\)( probability that a boy is chosen )

\(\frac{g}{20}\,=\,\frac23(\frac{b}{20})\\~\\ \frac{g}{20}\,=\,\frac23(\frac{20-g}{20})\\~\\ g\,=\,\frac23(20-g)\\~\\ \frac32g\,=\,20-g\\~\\ \frac32g+g\,=\,20\\~\\ \frac52g\,=\,20\\~\\ g\,=\,8\)

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The scale drawing of a rectangular yard measures (2x² + 2) by (x + 4). If the area of the scale drawing and the

area of the actual yard are in the ratio 12:140, find an expression for the area of the actual yard in expanded form.

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area of scale drawing  =  ( length )( width )  =  (2x² + 2)(x + 4)

area of scale drawing / area of actual yard  =  12 / 140

                                                                                        Substitute  (2x² + 2)(x + 4)  for  area of scale drawing

(2x² + 2)(x + 4) / area of actual yard  =  12 / 140            Now we just have to solve for  area of actual yard

                                                                                        Cross multiply

(140)(2x² + 2)(x + 4)  =  12(area of actual yard)

                                                                                        Divide both sides of the equation by  12

(140)(2x² + 2)(x + 4) / 12  =  area of actual yard

area of actual yard  =  (140)(2x² + 2)(x + 4) / 12

                                                                                        Expand the right side of the equation

area of actual yard  =  (140)(2x³ + 8x² + 2x + 8) / 12

area of actual yard  =  (280x³ + 1120x² + 280x + 1120) / 12

                                                                                                 Divide the numerator and denominator by  4

area of actual yard  =  (70x³ + 280x² + 70x + 280) / 3

number of tennis balls Sam has  =  14x + 3

number of tennis balls Jerri has  =  7x + 8

double the number of tennis balls Jerri has  =  2 * (7x + 8)

double the number of tennis balls Jerri has  =  14x + 16

14x + 16   is how many more than   14x + 3   ?

(14x + 16) - (14x + 3)  =  14x + 16 - 14x - 3  =  16 - 3  =  13

If Jerri doubles the number of tennis balls she has, she will have  13  more balls than Sam has.