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Arcs of inscibed angles cover twice as many degrees as the angle.

The smaller arc is 64 degrees

the larger LMN is 360 - 64 = 296 degrees

7) 

     x - 6                               x  -  6

   ___________   =         __________

  x^2 + 5x + 6                  (x + 2) ( x + 3)

We have no "holes" (one of the factors would have to cancel for this to happen )....the vertical asymptotes are at x = -3  and x = -2

8)  We have   a     same degree / same degree      situation....thus....the horizontal asymptote will be the ratio of the lead coefficients in each polynomial  =  -3/9   =  -1/3

Here's a pic

We have two areas.....one is a parallelogram  with a base of 6  and a height of  3

So.......its area  is :   base * height  =  6 * 3   =  18 units^2

We also have a triangle with the same base and a height of 2....so its area is :

(1/2) base * height  =  (1/2)(6) (2)  =  6 units^2

So.....the total area is   (18 + 6)   =  24 units^2

If  we have a root  at  (-3,0)  with a multiplicity of 2  and the point (2,0) is aslo on the graph.....we have three roots.....but this is a polynomial of degree 2, so it can't possibly have three roots.....did you make a mistake ??

Well....graphing is the fastest....however....the solution points could be difficult to interpret in some cases

But.....without a calculator or grapher, substitution is the fastest

(f/ g) (x)   =

   4x^2 + 6x

_____________

 2x^2 + 13x + 15

   2x(x + 6)

____________

(2x + 3) (x + 5)

Your suggestion will work only if the 5th term (-7) is wrong, which is quite likely because -17, which will work, is very close to -7 that he/she has written down.

Hi

if you still need help with this just leave a message here and i'll get back to you

cheers

There is no "best" method for factoring only what is preferred...or easiest to use. You decide which method works best for you! The one (or more) which gets you the correct answer!

The formula is a combination of both ie (-1)ⁿ (4n-3)  where n = 1,2,3,....

you see the (-1)ⁿ is just the mathematical way of putting a minus sign in front of a number.

odd powers give a minus and even powers give a plus eg (-1)³ = -1 ,  (-1)⁶ = 1

and of course the 4n-3 gives you  1,5,9,......

30 g  x  1 cal/gm-c  x  (80-40) c   x 4.184 j/cal = 5020.8 j

Second one

    .....first you have to warm the water....then convert it to steam

30 g x 4.186 j/gm-c x (100-50 c)    +   30 gm x 2260 j/gm   =  74079 j

The easiest thing to do in this situation is to draw a picture.

Then we label the information.

What we can do is use the 100 feet as a line. So I'm going to extend the bottom line to make it just under the top point. Then I'm going to move the bottom line down. The new angles are 60 and 30 degrees.

Now we can use SOH CAH TOA to find the hypotenuse. In this case, we would use cosine, since we have 100ft as the adjacent side and we're trying to find the hypotenuse.

The rest is using trig.

\(cos(30°)=\frac{100}{x}\)

\(\frac{\sqrt{3}}{2}=\frac{100}{x}\)

\(200=\sqrt{3}x\)

\(x=\frac{200}{\sqrt{3}}\)

\(x= 115.4701 ft\)

I read your question as:  ({e})^(-10/3 x) + ({e})^(-7/3 x) + ({e})^({-x}) = 0.15 , as Melody suggested in Equation (1).

"Mathematica 11 Home Edition" gives this answer for x, by iteration only. It does not give any step by step solution.

x = 1.97568 . This value of x does balance the equation as written above.

If it is Equation(2) as suggested by Melody, then there is no "real" solution for x. There is, however, a complex solution, and "Mathematica 11" gives this complex value for x:

x ≈ 0.651484 - 0.910470 i . But no explanation! Sorry about that.

First you have to warm the ice from -4 c to 0 c

74 gm * .5 cal/gm-c  * (4 c)  = 148 cal =   619.23 j         (1 cal = 4.184 j )

Then you have to melt the ice....this occurs at a constant temperature of 0 c

74 gm * 334 J/g  =   24716 j

Then you have to warm the water (which is now at 0 c) to 52 c

74 *1 * 52 = 3848 calories =16100 j

619.23 + 24716 + 16100 j = 41435.3 j =  41.4353 kj

tan 26.8 = opposite/ adjacent  =  opposite/100        solve for opposite to get HEIGHT  (50.51 ft)

then   sin 30 = opposite/hyp  to get hyp...the lenght of the track.    101.02 ft

Thank You, GA.

Since no one answered my last question I'm going to post it again.
I have 3 pieces of candy to place in 4 lunch boxes.
In how many ways can I do this if exactly two of the candies are the same (but the third is different)
and all of the lunch boxes are different?
Same means indistinguishable and different means distinguishable.

Let candi 1 = sort a
Let candi 2 = sort a
Let candi 3 = sort b

\(\begin{array}{|c|c|c|c|c|} \hline & \text{Box $1$} & \text{Box $2$} & \text{Box $3$} & \text{Box $4$} \\ \hline 1& b & a & a & \\ 2& b & a & & a \\ 3& b & & a & a \\ \hline 4& a & b & a & \\ 5& a & b & & a \\ 6& & b & a & a \\ \hline 7& a & a & b & \\ 8& a & & b & a \\ 9& & a & b & a \\ \hline 10& a & a & & b \\ 11& a & & a & b \\ 12& & a & a & b \\ \hline \\ \hline 13& ba & a & & \\ 14& ba & & a & \\ 15& ba & & & a \\ \hline 16& a & ba & & \\ 17& & ba & a & \\ 18& & ba & & a \\ \hline 19& a & & ba & \\ 20& & a & ba & \\ 21& & & ba & a \\ \hline 22& a & & & ba \\ 23& & a & & ba \\ 24& & & a & ba \\ \hline \\ \hline 25& baa & & & \\ 26& & baa & & \\ 27& & & baa & \\ 28& & & & baa \\ \hline \\ \hline 29& aa & b & & \\ 30& aa & & b & \\ 31& aa & & & b \\ \hline 32& b & aa & & \\ 33& & aa & b & \\ 34& & aa & & b \\ \hline 35& b & & aa & \\ 36& & b & aa & \\ 37& & & aa & b \\ \hline 38& b & & & aa \\ 39& & b & & aa \\ 40& & & b & aa \\ \hline \end{array}\)

40 ways

5/2

No time to post a full solution at the moment.

X = t + 1 , y = 2 - 32t.  ....not sure about the speed

Heureka, your answer is correct if a box is restricted to a maximum of one piece of candy.

nCr(4, 1) +  nCr(3, 2) = 12

Reasoning:

There are nCr(4, 1)  =  4 ways to place the unique piece of candy in one of the four (4) boxes.  Then the 2 indistinguishable candies can be placed in the other 3 boxes nCr(3, 2)  = 8 ways.  There are (4+8) = 12 ways to distribute the candies.

If there are no restrictions on the maximum then:

\(\hspace{15pt} \left( \dfrac{4!}{2!}\right)* \left(2 \right) \;+\; \left( \dfrac{4!}{2!}\right) \;+\; \left(4\right)\\ \text {Reasoning:}\\ \text {All candy in 1 of 4 boxes (4 ways)}\\ \text {2 combinations (XX|Y & XY|X) distributed to boxes in (2)(4*3) = (2)(4!/2!) =(24) ways;}\\ \text {and 1 candy to each box distributed to boxes in (1)(4*3*2)/2 = 4!/2! = (12) ways.} \\ \text {Total 4+24+12 = 40 ways.} \)

Are you sure you have copied it correctly?

Do you mean   

\({e}^{(-10/3)x}+{e}^{(-7/3)x}+{e}^{-x}=0.15\qquad(1)\\~\\ or\\~\\ {e}^{-10/(3x)}+{e}^{-7/(3x)}+{e}^{-x}=0.15\qquad (2)\)

.453 moles O2

g(x) = 3 + 2sin(x)

i have put the question but I want the steps

The function of f and g are such that

f(x) = 5x + 3 

g(x) = ax + b

g(3) = 20 and f^-1(33) = g(1)

Find the value of a and b 

1.

\(\begin{array}{|lrcll|} \hline & g(x) &=& ax+b \\\\ & g(3) &=& a\cdot 3+b \quad & | \quad g(3) = 20 \\ & 20 &=& a\cdot 3+b \\ \mathbf{(1)} & \mathbf{3a+b} & \mathbf{=} & \mathbf{20} \\ \hline \end{array}\)

2.

\(\begin{array}{|lrcll|} \hline & f(x) &=& 5x+3 \\ & y &=& 5x + 3 \\ & 5x &=& y-3 \\ & x &=& \frac{y-3}{5} \quad & | \quad \text{change $x$ and $y$ } \\ & y &=& \frac{x-3}{5} \\ & f^{-1}(x) &=& \frac{x-3}{5} \\\\ & f^{-1}(33) &=& \frac{33-3}{5} \\ & &=& \frac{30}{5} \\ & f^{-1}(33) &=& 6 \quad & | \quad f^{-1}(33) = g(1) \\ & g(1) &=& 6 \quad & | \quad g(1) = a\cdot 1 + b \\ & a\cdot 1 + b &=& 6 \\ \mathbf{(2)} & \mathbf{a+b} & \mathbf{=} & \mathbf{6} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \mathbf{(1)} & \mathbf{3a+b} & \mathbf{=} & \mathbf{20} \\ \mathbf{(2)} & \mathbf{a+b} & \mathbf{=} & \mathbf{6} \\ \\ \hline (1)-(2) & 3a+b -a -b &=& 20-6 \\ & 2a &=& 14 \quad & | \quad : 2 \\ & \mathbf{a} &\mathbf{=}& \mathbf{7} \\\\ & a+b &=& 6 \quad & | \quad a=7 \\ & 7+b &=& 6 \\ & b &=& 6-7 \\ & \mathbf{b} &\mathbf{=}& \mathbf{-1} \\ \hline \end{array}\)

a = 7 and  b = -1