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I may have misread your question!! I multiplied the constant k, by the first term only. Looking at it more closely, it appears that it should be multiplied by the whole expression as follows:

∑k*[(ln(k)/0.34657359) - (ln(k)/0.34657359)], for k=1 to 1000

If this is accurate, then N = 0 !!!.

Let   \(f(x)=\left(\frac37\right)^x\)     be a function that is defined on the domain \([0,\infty)\). Find the range of the function.

f(0)=1     f(infinity)=+0      all the other values in the given domain are between these  so

range (0,1]

strange...ask melody or another admin

∑[(k*(ln(k)/0.34657359)) - (ln(k)/0.34657359)], for k=1 to 1000

N≈9.23734613867.....×10^6

Note: I used natural log(ln) instead of common log(base 10), but it should not alter the result.

Also I converted log(sqrt(2)) to a constant of 0.34657359. Check my result by summing up 3-4 terms using the notation above. I did so and it seems OK.

1. In triangle ABC, AB=12 units and AC=9 units. Point D is on segment BC so that BD:DC=2:1. If AD=6 units, what is the length of segment BC? Express your answer in simplest radical form. 

Let BC = x

cos rule:

\(\begin{array}{|lrcll|} \hline & 6^2 &=& 9^2+(\frac13 x)^2 - 2\cdot 9 \cdot \frac 13 x \cos( C ) \\ & 6^2 &=& 9^2+(\frac13 x)^2 - 6 x \cos( C ) \\ & 6 x \cos( C ) &=& 9^2-6^2+\frac{x^2}{9} \\ (1) & 6 x \cos( C ) &=& 45+\frac{x^2}{9} \\ \hline \end{array}\)

cos rule:

\(\begin{array}{|lrcll|} \hline & 12^2 &=& 9^2+x^2 - 2\cdot 9 \cdot x \cos( C ) \\ & 12^2 &=& 9^2+x^2 - 18 x \cos( C ) \\ & 18 x \cos( C ) &=& 9^2-12^2+x^2 \\ & 18 x \cos( C ) &=& x^2-63 \quad & | \quad : 3 \\ (2) & 6 x \cos( C ) &=& \frac{x^2}{3}-21 \\ \hline \end{array} \)

(2) = (1):

\(\begin{array}{|rcll|} \hline 6 x \cos( C ) = \frac{x^2}{3}-21 &=& 45+\frac{x^2}{9} \\ \frac{x^2}{3}-21 &=& 45+\frac{x^2}{9} \\ x^2(\frac13-\frac19) &=& 66 \\ x^2(\frac{9-3}{27}) &=& 66 \\ x^2(\frac{6}{27}) &=& 66 \\ x^2 &=& 66\cdot (\frac{27}{6}) \\ x^2 &=& 11\cdot 27 \\ x^2 &=& 11\cdot 3^2 \cdot 3 \\ \mathbf{ x} & \mathbf{=} & \mathbf{3\sqrt{33}} \\ \hline \end{array}\)

The length of segment BC is \(\mathbf{3\sqrt{33}}\)

A square $DEFG$ varies inside equilateral triangle $ABC,$ so that $E$ always lies on side $\overline{AB},$ $F$ always lies on side $\overline{BC},$ and $G$ always lies on side $\overline{AC}.$ The point $D$ starts on side $\overline{AB},$ and ends on side $\overline{AC}.$ The diagram below shows the initial position of square $DEFG,$ an intermediate position, and the final position.

Show that as square $DEFG$ varies, the height of point $D$ above $\overline{BC}$ remains constant.

The LaTex isn't displaying clearly for me...if you  have (3/7)^x   as the equation...

at x = 0 the value of f(x) = 1    (i.e. y=1)      as x becomes larger and larger the equation becomes smaller and smaller.....reaching essentially y=0

So for the domain x = 0 to infinity the range is 1 to 0   .

Here is a graph:  (disregard portion of the graph left of y-axis)

\(f(x)=\sqrt{-6x^2+11x-4}\)

The domain is all real values of  x  such that

-6x² + 11x - 4  ≥  0

To solve this inequality, let's first solve this equation...

-6x² + 11x - 4  =  0       Using the quadratic formula...

x  =  \({-11 \pm \sqrt{11^2-4(-6)(-4)} \over 2(-6)}\,=\,{-11 \pm 5 \over -12}\)

x  =  \(\frac12\)         or        x  =  \(\frac{4}{3}\)

Since the x values that make  -6x² + 11x - 4  equal  0  are  \(\frac12\)   and   \(\frac{4}{3}\) ,

(and since  -6x² + 11x - 4  is continuous....)

the x values that make  -6x² + 11x - 4  greater than or equal to  0  will be either be...

those between \(\frac12\)  and  \(\frac{4}{3}\)  ,  that is,  those in the interval  \([\frac12,\frac43]\)

OR

those outside of that interval, that is, those in the interval  \((-\infty,\frac12]\cup[\frac43,\infty)\)

To determine which...

Let's test an  x  value in the interval  \([\frac12,\frac43]\) .  Does  x = 1  make  -6x² + 11x - 4  ≥  0    ?

-6(1)² + 11(1) - 4  ≥  0

1 ≥ 0      true

Let's test an  x  value in the interval  \((-\infty,\frac12]\cup[\frac43,\infty)\) . Does  x = 0  make  -6x² + 11x - 4  ≥  0 ?

-6(0)² + 11(0) - 4  ≥  0

-4 ≥ 0     false

So the  x  values that make  -6x² + 11x - 4  ≥  0  are those in the interval  \([\frac12,\frac43]\)

The domain of  f(x)  is  \([\frac12,\frac43]\)

Remember that  we cant have any values that make a denominator  =  0

So.... k cannot   =3    because that makes the denominator 0

I don't know where they get  -5/2......this functiion is certainly defined at k  = -5/2

See the graph, here : https://www.desmos.com/calculator/hh4osmk1fc

I've substituted x for k, but the idea is still the same....

For the first function

x^2 - 5x - 64  =2

x^2 - 5x -66 = 0    factor

(x - 11)  ( x + 6)  = 0

The only solution  for the first restriction  is  x  = -6  ⇒ (-6, 2)

For the second function

x^2 + 3x - 38  = 2

x^2 + 3x - 40  =0\

(x + 8) ( x - 5)  = 0

The only solution for the second restriction is  x = 5 ⇒  (5 , 2)

So....two points

This area forms a rectangle with vertices  (0,0), (1.5,1.5), (-1,4)  and (-2.5, 2.5)

The length of one side is  √ [2 * 1.5^2 ] =  1.5√2

And the length of the other side is √ [2 *2.5^2 ]  = 2.5√2

So  the total area  = 1.5√2 * 2.5√2  =  3.75 * 2    =  7 units^2

Here's a pic of the region :  https://www.desmos.com/calculator/zlpwagi3cx

i think so, it's the same for the previous answer, but just the inverse square model (1/4).

no need for sphere areas.

r u sure?

a)  \(\frac{3x}{4}-\frac{x}{8}\)

                             Multiply the first fraction by  \(\frac22\)

\(=\,\frac22\cdot\frac{3x}{4}-\frac{x}{8}\\~\\ =\,\frac{6x}{8}-\frac{x}{8}\)

                             Now that we have a commond denominator we can combine the terms.

\(=\,\frac{6x-x}{8}\\~\\ =\,\frac{5x}{8}\)

b)  \(\frac{3x}{4}-\frac{x}{3}\)

                                   Multiply the first fraction by \(\frac33\)  and the second fraction by  \(\frac44\)

\(=\,\frac33\cdot\frac{3x}{4}-\frac{x}{3}\cdot\frac44\\~\\ =\,\frac{9x}{12}-\frac{4x}{12}\\~\\ =\,\frac{9x-4x}{12}\\~\\ =\,\frac{5x}{12}\)

c)  Technically what you've written is  \(\frac{4}{15}x-\frac{3}{30}x\)    but I will do  \(\frac{4}{15x}-\frac{3}{30x}\)

\(\frac{4}{15x}-\frac{3}{30x}\)

                                  Multiply the first fraction by  \(\frac22\)

\(=\,\frac22\cdot\frac{4}{15x}-\frac{3}{30x} \\~\\ =\,\frac{8}{30x}-\frac{3}{30x}\\~\\ =\,\frac{8-3}{30x}\\~\\ =\,\frac{5}{30x}\)

                                  Now we can reduce this fraction by  5 .

\(=\,\frac{1}{6x}\)

read up on attenuation laws.

use inverse cube model for twice the distance. 4 x 1/8 = 0.5uW/m2

Circle N is a dilation of circle M with a scale factor of 3, so the radius of N is 3 times the radius of M.

The radius of circle N  =  3 * 12  =  36      so statement  A  is false.

Circle N is a translation of circle M, 1 unit left. So the center of N is one unit left of the center of M.

The center of circle M  =  one unit left of  (1, 10)  =  (1-1, 10)  =  (0, 10)      so statement  B  is true.

None of the transformations cause a loss of similarity, so statement  C  is true.

Circle N is a dilation of circle M with a scale factor of 3, so the diameter of N is 3 times the diameter of M.  So statement  D  is false.

Nice one, hectictar   !!!

2. If f(x)=ax^4-bx^2+x+5 and f(-3)=2, then what is the value of f(3)?

If  f(-3)  = 2, we have

2  = a(-3)^4 - b(-3)^2 - 3 + 5

2  = 81a - 9b + 2

0 = 81a - 9b  

9b  = 81a

b = 9a

f(3)  =  a(3)^4 - (9a)(3)^2 + 3 + 5

f(3)  = 81a - 81a  + 3 + 5

f(3)  = 8

Simplify the following:

(b^2 + 2 b - 24)/(2 b^2 - 72)

The factors of -24 that sum to 2 are 6 and -4. So, b^2 + 2 b - 24 = (b + 6) (b - 4):

((b + 6) (b - 4))/(2 b^2 - 72)

Factor 2 out of 2 b^2 - 72:

((b + 6) (b - 4))/(2 (b^2 - 36))

b^2 - 36 = b^2 - 6^2:

((b + 6) (b - 4))/(2 (b^2 - 6^2))

Factor the difference of two squares. b^2 - 6^2 = (b - 6) (b + 6):

((b + 6) (b - 4))/(2 (b - 6) (b + 6))

((b + 6) (b - 4))/(2 (b - 6) (b + 6)) = (b + 6)/(b + 6)×(b - 4)/(2 (b - 6)) = (b - 4)/(2 (b - 6)):

(b - 4) / (2(b - 6))

 \( f(x) = \sqrt{\dfrac{x+1}{x-1}}\qquad\qquad\)

\(\qquad\qquad g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}.\)

If these functions are the same, they have the same domain.....

Notice that the domain of the first function is   (-inf, -1) U (1, inf )

But  the domain of the second function is only (1, inf)......any  negative in the denominator makes this function undefined for the set of real numbers.....

So....these functions are not the same

1.   \(f(x)=\sqrt{3-\sqrt{5-\sqrt{x}}}\)

\(\sqrt{x}\)   is part of the function so the domain has the restriction:

x ≥ 0

\(\sqrt{5-\sqrt{x}}\)   is part of the function so the domain has the restriction:

\(5-\sqrt{x}\geq0\\ 5\geq\sqrt{x}\\ 25\geq x\\ \mathbf{x\leq 25}\)

\(\sqrt{3-\sqrt{5-\sqrt x}}\)   is part of the function so the domain has the restriction:

\(3-\sqrt{5-\sqrt x}\geq0\\ 3\geq\sqrt{5-\sqrt x}\\ 9\geq5-\sqrt x\\ 9+\sqrt x\geq5\)

\(\sqrt x\geq-4\)          This is true for all real numbers.

The domain restrictions are:  x ≥ 0  and  x ≤ 25

So the domain is just

0 ≤ x ≤ 25

Here is the diagram

sin of an angle   =  length of opposite side / length of hypotenuse

sin 60°   =   y / 8

                             Multiply both sides of the equation by  8 .

8 sin 60°  =  y

y  =  8 sin 60°

                             sin 60°  =  √3 / 2

y  =  8( √3 / 2 )

y  =  4√3

\(\frac{3g}{g^3-9g}\)

To find the non-permissible value of  g , set the denominator equal to zero.

g³ - 9g   =  0    To solve this equation for  g , we need to factor the left side.

                        Start by factoring a  g  out of both terms.

g(g² - 9)  =  0   Now we can factor  g² - 9  as a difference of squares.   g² - 9  =  (g + 3)(g - 3)

g(g + 3)(g - 3)  =  0      Set each factor equal to zero.

g  =  0     or     g + 3  =  0     or     g - 3  =  0

                          g  =  -3                 g  =  3

The non-permissible values of  g  are  0 ,  -3 , and  3

Nonpermissable values would result in the denominator = 0

so as guest found   3 and -3  are two possibilities   as is 0