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The original function is

f(x)  = 2x  - 1

So

f(-x)  means  that  we replace  "x"  with "-x"  in the original function

So

f(-x)  = 2(-x)  - 1  =   -2x - 1

Here are both graphs....:  

https://www.desmos.com/calculator/ob8iw0ihx3 1

Note  the  effect of replacing "x" with "-x"    .....it just "reflects" the original graph across the y axis.....!!!

I think heureka answered this here :

https://web2.0calc.com/questions/correct#r2

We need to write the equation of a plane, here, Julius.....

Since the prism is bisected  by the xy plane,

Let D  = (0, 0 ,-3 )

Let  A  = (0, 0, 3)

Let E  = (4, 0,3)

Let G   =(4,0,-3)

We can form the following  vectors  DG  = (4,0, 0)    and  DE  = (4, 0, 6)

[These aren't the only two possibilities......we just need any two vectors formed by the points]

Note that all the points lie in the xz  plane....so...we need a vector that is orthagonal to this plane

The cross-product of the two vectors will be orthagonal to both of these vectors...so we have

                         i          j           k       i         j

DG X DE   =    4         0          0       4        0

                        4        0           6       4        0       

 ( [ 0 * 6]i   + [ 0 * 4 ] j  + [4 * 0] k   -  (  [ 4 * 0 ] k  + [ 0 *0] i  +  [ 6 * 4]j )   =

0i   + 0 j   +  0 k   - 0 k - 0i  - 24 j

So....we have

0i  -24j - 0k       =  { 0, -24, 0 }    this is the vector  that is orthagonal to both vectors ....[and hence,  the plane containing those vectors]

And  the equation of the plane becomes

0 (x - 4) -24 (y - 0)  + 0 (z - 3)   =  0

-24y    =  0 

4.  We can use the secant-tangent theorem to solve this, ACG...

We have something like this  (not to scale) :

PA * PB  = PT^2

PA * PB  = (AB - PA)^2

PA * ( AB + PA)  = AB^2 - 2PA*AB + PA^2

PA*AB + PA^2 = AB^2  - 2PA*AB + PA^2

3*AB + 3^2  = AB^2  - 2*3*AB  + 3^2

3AB + 9  = AB^2 - 6AB + 9

3AB  = AB^2 - 6AB     rearrange as

AB^2  - 9AB  = 0       factor

AB( AB - 9)  =  0

Seting each factor to 0 and solving for AB  we have that either

AB  = 0  (reject)    or  AB  = 9   (accept)

So.....PB  = AB + PA

PB  = 9 + 3

PB  = 12

There is a formula for the sum of the divisors, but you have to know the number whose divisors are to be summed up. For your particular problem, it goes like this:

[(2^(i + 1) - 1) / (2 - 1)] * [(3^(j + 1) - 1)] / (3 - 1) = 600

I don't know how to formally solve for 2 variables in one equation without resorting to trial and error, which quickly leads to i = 3 and j = 3, which give: 2^3 x 3 ^3 = 216. Then you would find the divisors of 216, as done above, which sum up to 600.

2^3 x 3^3 =216

Divisors of 216 =1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 27 + 36 + 54 + 72 +108 + 216 = 600

i + j = 3 + 3 = 6

To get f^-1 rewrite f as:

   \(f=\frac{a}{x+2}\\x+2=\frac{a}{f}\\x=\frac{a}{f}-2\)

Now change f into x and x into f^-1:

\(f^{-1}=\frac{a}{x}-2\)

We have: \(f(0)=\frac{a}{2}\text{ and }f^{-1}(3a)=\frac{a}{3a}-2\rightarrow-\frac{5}{3}\)

I'm sure you can take it from here!

I think it's Ellipse. 

Thanks so much, I was looking desperatley for a way to do it without trial and error, but with two variables and one equation I guess that's unavoidable.

According to the laws of geometry, 

Angle B is one half of arc AC

\((4x-3.5)\cdot2=4x+17\)

\(8x-7=4x+17\)

\(4x=24\)

\(x=6\)

Plugging this into 4x + 17, we get:

\(4\cdot6+17=\boxed{41}\)

I hope this helped,

Gavin

Hi Guest!

We can rewrite the left side of the equation,

using the arithmetic series sum formula:

\(S_n=\frac{n(a_1+a_n)}{2}\)

We get:

\(100=\frac{n(2a+n-1)}{2}\)

\(200=n(2a+n-1)\)

Hence, n must be a factor of 200.

 \( \begin{array}{c|c|c} n & 2a + n - 1 & a \\ \hline 2 & 100 & 99/2 \\ 4 & 50 & 47/2 \\ 5 & 40 & 18 \\ 8 & 25 & 9 \\ 10 & 20 & 11/2 \\ 20 & 10 & -9/2 \\ 25 & 8 & -8 \\ 40 & 5 & -17 \\ 50 & 4 & -45/2 \\ 100 & 2 & -97/2 \\ 200 & 1 & -99 \end{array} \ \)

Since a must be a positive integer, the solutions are \((a,n) = \boxed{(18,5)\text{ and }(9,8)}\). 

I hope this helped,

Gavin

These are homework problems! Should be attempted first!

Looking at the triangle on the left, we can write its base as 50 - x

And, using the Pythagorean Theorem , we can solve this :

(50 - x)^2  + 30^2  =  x^2       simplify

x^2 - 100x + 2500 + 900 = x^2       subtract x^2  from both sides

-100x  +  3400  =  0

3400 = 100x       divide both sides by 100

34  =  x  =  AB

7 / sin(80) = 5 / sin(J)

7 / 0.9848 = 5 / sin(J)   cross multiply

7*sin(J) =0.9848*5

7*sin(J) =4.924             divide both sides by 7

sin(J) =4.924 / 7

J = Arcsin(4.924 /7) =44.70 = ~45 degrees.

I can barely see your questions! Try and make them a bit larger next time.

15)

Area= 1/2 x 13^2 x sin(60) =73.2 in^2

16)

Tan(x) = 86

x =Arctan(86) =89.33 degrees.

17)

Cos(x) =7/14 = 1/2

x =Arccos(1/2) =60 degrees

18)

Sin(x) =2 / 25

x =Arcsin(2/25) =4.59 degrees.

2) Area of the isosceles triangle ABC =12 [By bisecting BC, we have, 3, 4, 5 right triangle]

Since O is the circumcenter of triangle ABC, then:

The circumradius OC=OB=3.125. The circumradius R = a.b.c / 4*Area =5.5.6 / 4*12=3.125.

Then the area of the isosceles triangle OBC with sides OC=OB=3.125 and BC=6 and h=0.875

=2 5/8 units^2

Correct?

\(\text{The unit vector in the opposite direction of $ \vec{u}$ is $ \dfrac{-\vec{u}}{|\vec{u}|}=-\dfrac{\vec{u}}{7} $ }\)

How would you write the equation, y²+8x=0 rotated 30° in general form?

\(\frac{1}{4}x^2-\frac{\sqrt{3}}{2}xy+\frac{3}{4}y^2+4\sqrt{3}x+4y=0\)

As Guest#6 says, there is actually a Theorem called the "Chicken McNugget Theorem" and is discussed in some significant detail here:

https://artofproblemsolving.com/wiki/index.php?title=Chicken_McNugget_Theorem

I agree with Guest #6, we are looking for 7x + 10y + 53z = d where d is the sum required and x, y and z are all positive.

That means that most of the smaller sums are not possible, (we are not allowed to pay 8 units for example, by tendering 24 7 unit coins and receiving 16 10 ' s in change).

The answer can be arrived at by what might be regarded as trial and error. 

Forget the 53 for the moment and just look at the 7 and 10 unit coins, and prior to that the various multiples of 7,

7, 14, 21, 28, 35, 42 ,49, 56, 63, and notice that that the units digits cover each of the numbers from 1 to 9.

Some examples.

Suppose the sum required were 89, to make this up, look at the final digit 9 and that will lead to 89 = 49 + 40 = 7*7 + 4*10.

Similarly 74 = 14 + 60 = 2*7 + 6*10,

68 = 28 + 40 = 4*7 + 4*10, etc.

If the sum is large we may or maynot choose to make use of the 53 unit coins.

For example if the sum required was say 732, we could still use just 7's and 10's, 732 = 42 + 690 = 6*7 + 69*10, (there's nothing in the question to say that we can't do that), or we could say that 732 = 636 + 96 = 12*53 + 8*7 + 4*10.

Going back to smaller numbers, we couldn't make up a sum of 39, for example, because, from the 7's multiples we would have to use the 49 and that is too big, it's greater than the required 39.

Looking at it this way the largest number that can't be made up, (with one exception), is 46, as it requires the use of 56.

All numbers above this, with one exception, can be catered for.

47 = 7 + 40, 48 = 28 + 20, 49 = 49, 50 = 50, 51 = 21 + 30, 52 = 42 + 10, 53 = ????.

53 doesn't work, the 3 means that we have to use 63 which is too big,  and that's why 53 is included in the question as a third denomination coin.  

There is some background analysis using the Euclidean algorithm, it leads to

\(\displaystyle \frac{2S}{7}\leq k \leq\frac{3S}{10}\quad ,\)

where S is the sum required and k is a parameter.

For a given value of S, it's necessary to be able to find an integer value for k in order that the sum can be made up with 7's and 10's.

If, for example S = 46 we have 92/7 (=13.14..)<= k <= 138/10 (=13.8), so no integer k exists, so 46 can't be catered for,

while, for example,

if S = 47, 94/7 (=13.43..) <= k <= 141/10 (=14.1), so k = 14 and a solution can be found.

Tiggsy

The midpoint  of this segment is  ( 3, 5)

And the slope between the two segment endpoints  is  [7 - 3] / [ 5 - 1]   = 4/4  = 1

So.......the perpedicular bisector  will have a negative reciprocal slope  = -1

And the equation of this bisector is given  by :

y = -1(x - 3) + 5

y = -x + 3 + 5

y = -x + 8

So

x + y  =  8     

And  "b"  =  8

The graph of the line $x+y=b$ is a perpendicular bisector of the line segment from $(1,3)$ to $(5,7)$.

What is the value of b?

\(\begin{array}{|lrcll|} \hline & \vec{x} &=& \dbinom{ \frac{1+5}{2} } { \frac{3+7}{2} } + \lambda \binom{ -(7-3) } {5-1 }\\ & &=& \dbinom{3} { 5 } + \lambda \binom{ -4 } { 4 }\\ & \dbinom{x}{y} &=& \dbinom{3-4\lambda}{5+4\lambda} \\\\ (1) & x &=& 3-4\lambda \\ (2) & y &=& 5+4\lambda \\\\ (1)+(2): & x+y &=& 3-4\lambda +5+4\lambda \\ & x+y &=& 3 +5 \\ & x+y &=& 8 \quad & | \quad x+y=b\\ & &&& | \quad \mathbf{b = 8} \\ \hline \end{array}\)