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Regular price=P
8P + 4*$1 =$120
8P = $116
P = $116 /8
P=$14.50 - regular price of one shirt.

The number is = 2000 - zero is cosidered and even digit. The next one without zero would be 2222.

Let x = x^4

x^2 - 12x-64

(x-16)(x+4)                      put x^4 back in

(x^4-16)(x^4+4)               let x = x^2

(x^2-16)(x^2+4)

(x-4)(x+4)(x^2+4)             put x^2 back in

(x^2-4)((x^2+4)(x^4+4)

(x-2)(x+2)(x^2+4)(x^4+4)    

(x-2)(x+2)(x^2+4)(x^2+2x+2)(x^2-2x+2)       I think this is as far as I can go with it....anyone else? 

To find the primary fourth root of the complex number  a + b·i,  place the number into  r·cis( theta ) form.

r  =  sqrt( a² + b² )

theta  =  tan^-1( b/a )    [make certain that the angle is in the correct quadrant; in this case, the 2nd[

The primary fourth root will be:  r^1/4·cis( theta/4 ).

To find the other three roots, change the angle by adding  pi/2, pi, and 3pi/2 to the original angle.  

123 , 234 , 345 , 456 , Total =  4

Probability = 4 / 6^3 = 4 / 216 = 1 / 54

A    Graph is shifted so that the value  at 'x' is now the value that was at x-3

         so    f(x-3)

B    Graph is now showing the value at 'x' that WAS the value at -x

         sooo   g(-x)

I know but give answer or bigger hint

Hint: You can use a Venn diagram.

Another way:   (doing a lot of factoring)

Numerator:  100! + 99!  =  99!(100 + 1)  =  99! · 101

                        98! + 97!  =  97!(98 + 1)  =  97! · 99

                        96! + 95!  =  95!(96 + 1)  =  95! · 97

                                  ...

                                2! + 1!  =  1!(2 + 1)  =  1! · 3

Total numerator:  99! · 101 · 97! · 99 · 95! · 97 · ... · 1! ·3

Denominator:  100! - 99!  =  99!(100 - 1)  =  99! · 99

                            98! - 97!  =  97!(98 - 1)  =  97! · 97

                            96! - 95!  =  95!(96 - 1)  =  95! · 95

                                     ...

                                  2! - 1!  =  1!(2 - 1)  =  1! · 1

Total denominator:  99! · 99 · 97! · 97 · 95! · 95 · 1! · 1

Notice that all the factorials of the numerator and the denominator cancel each other.

Notice that all the single numbers cancel each other except for the 101 in the numerator

and th 1 in the denominator.

Answer  =  101 / 1  =  101

x = sqrt 30 cos 110

y = sqrt 30 sin 110 .......

First, you have to sum them separately using "closed form" formulas:

The top one:= product_(n=1)^50((101 - 2 n)! + (-2 (-51 + n))!) =7.733930684 E+3510

The bottom one: = product_(n=1)^50(-(101 - 2 n)! + (-2 (-51 + n))!) = 7.657357113 E+3508

7.733930684 E+3510 / 7.657357113 E+3508 =101 - The result

PA and PB are tangents to circle(O).  Draw PO. Draw AB. Let X be the point where PO and aB intersect.

PA = 12   and   OA = 9

Because triangle(OPA) is a right triangle, PO = 15.

Triangle(AXO is similar to triangle(PAO)   --->   AX / AO  =  PA / PO   --->   AX / 9  =  12 / 15

--->   AX  =  7.2

Since X is the midpoint of AB   --->   AB  =  14.4.

There are two externally tangent circles, circle(A) with radius = 10 and circle(B) with radius = 7.

Therefore, AB = 17.

There is an externally tangent line (with both circle(A) and circle(B) on the same side of the line)

XY, where X is the point of intersection with circle(A) and Y is the point of intersection with circle(B).

Therefore, AX = 10 and BY = 7.

Extend both AB and XY so they intersect at point C.

Triangle(ACX) will be similar to triangle(BCY) because they have corresponding congruent angles.

Call BC = x.  This makes AC = x + 17.

This means that:  AX / BY  =  AC / BC.   --->   10 / 7  =  (x + 17) / x

--->   10x  =  7(x + 17)   --->   10x  =  7x + 119   --->   3x  =  119   --->   x  =  119/3

BC  =  119/3

If I understand your question, then you have a set consisting of 15 terms of an arithmetic sequence:

[46 - 4] / 3 + 1 =15 terms in your set.

To choose 3 distinct number from the set:

You have: 15C3 =455 ways of summing up 3 distinct numbers from your set of 15 numbers.

1)  The line  x + 2y + 4  =  0   --->   2y  =  -x - 4   --->   y  =  -½x - 2   --->   m  =  -½.

      Since the line  6x - ny + 5  =  0  is perpendicular to the other line, its slope must be  +2.

      6x - ny + 5  =  0   --->   -ny  =  -6x - 5   --->   ny  =  6x + 5   --->   n  =  (6/n)x + 5/n

      The slope of this line is  6/n.

      This slope must equal +2   --->   6/n  =  2   --->   6  =  2n   --->   n  =  3

2)  x + 2y  =  10  has these intercepts  (0,5)  and  (10,0).

     The line with an infinite number of x-intercepts is the x-axis.

     The line with an infinite number of y-intercepts is the y-axis.

      The vertices of the triangle are  (0,0),  (0,5),  and  (10,0).

      The area  =  ½·5·10  =  25.

Draw the regular hexagon and the inscribed circle.

Draw a triangle from the center of the circle, along a radius to a point of tangency of the circle and the 

hexagon, and along the tangent to a corner.

This triangle is a 30^o-60^o-90^o triangle with the length of the side from the point of tangency to the

corner = 10.

This will make the radius = 10·sqrt(3).

For a circumscribed quadrilateral, the sum of a pair of opposite sides must equal the sum of the other pair.

Therefore:  6 + ?  =  5 + 13  or  ? = 7.

This follows from the theorem that the two tangents draw from a point outside the circle are congruent.

#1 tells me it’s a city. Nepal is the only city. The others are countries so…

The first problem:  log₅(8x) + 2log₅(x)  =  3

--->   log₅(8x) + log₅(x²)  =  3        (property of logs: a multiplier goes in as an exponent)

--->              log₅(8x ·x²)  =  3        (property of logs: adding logs = multiplying numbers)

--->                   log₅(8x³⁾ = 3         (simplify)

--->                         8x³  =  5³        (rewriting in exponential form)

--->                          x³  =  5³/8       (divide)

--->                          x³  =  5³/2³      (rewrite 8 as 2³)

--->                            x  =  5/2        (find the cube root)

To solve     2cos²(x) - sin(x) - 1  =  0     use the identity:  cos²(x)  =  1 - sin²(x):

--->   2[ 1 - sin²(x) ] - sin(x) - 1  =  0

             2 - 2sin²(x) - sin(x) - 1  =  0

                -2sin²(x) - sin(x) + 1  =  0

                  2sin²(x) + sin(x) - 1  =  0

          [ 2sin(x) - 1 ][ sin(x) + 1 ]  =  0

So:  either  2sin(x) - 1  =  0   --->   2sin(x)  =  1   --->   sin(x)  =  ½   --->   x  =  sin^-1(½)

              or  sin(x) + 1  =  0   --->     sin(x)  =  -1   --->                                x  =  sin^-1(-1)

From here, you can get the answers ...

Ah okay, small detail that I missed. Thanks so much!

The problem arose when you set  n² - 2  =  14   --->   n²  =  16

Normally, you would get two answers,  -4  and  4.

However, in this case there is a restriction upon using  n² - 2 

---   you are only supposed to use  n² - 2  when  n < 0

---   so, for this case, the answer is only -4  (you need to exclude the answer 4)

Everything else that you did was "textbook perfect" ...

cause a constant term is just an integer, and there isn't a constant term here

the answer is actually 0 since there is no constant term