bingboy

There are 2 ways to start with the letter A: either from the first A in the array, or from the second A. To reach the second A, we must move down once. Once we have reached the first or second A, we can move in any direction to reach the letter R. There are 3 ways to do this: we can move down, right, or down and then right. Once we have reached the letter R, we can again move in any direction to reach the letter C. There are 3 ways to do this: we can move down, left, or left and then down. Once we have reached the letter C, the last step is to move up to reach the letter H, and there is only 1 way to do this. Therefore there are 2×3×3×1=18​ ways to spell the word "ARCH" in the array provided.

Virginia writes down the value of 100 products of integers from 0 to 9. For each product, she has 10 choices for the first integer and 10 choices for the second integer, for a total of 10⋅10=100 possibilities. However, some of these products will be the same.

For example, both 2⋅5 and 5⋅2 are equal to 10, so these two products are counted twice. To avoid this overcounting, we can use the following formula to count the number of distinct products:

(rn+r−1​)

where n is the number of distinct elements in the set and r is the number of elements in each product. In this case, we have n=10 distinct elements (the digits 0 to 9) and r=2 elements in each product. So, the number of distinct products is:

(210+2−1​)=(211​)

Using the combination formula, we can calculate that there are (211​)=55 distinct products. Therefore, Virginia writes down 55​ distinct numbers.

The function f(x)=x+81​+x−8​1​+8−x​1​ is undefined when any of the denominators are zero. The first denominator, x+8, is zero when x=−8. The second denominator, x−8​, is zero when x=8. The third denominator, 8−x​, is zero when x=8. Therefore, the function is undefined when x=−8 or x=8.

We also need to consider the restrictions imposed by the square root functions. The square root of a number is undefined when the number is negative. Therefore, the second denominator, x−8​, is undefined when x−8<0, which is when x<8. Similarly, the third denominator, 8−x​, is undefined when 8−x<0, which is when x>8.

Putting all of this together, the function f(x) is undefined when x=−8, x=8, x<8, or x>8. Therefore, the domain of f(x) is the union of the intervals (−∞,−8)∪(8,∞)​.

There is a nice soluiton here: https://web2.0calc.com/questions/hxlp

Since DI =3, BD = 4  and BI = 5, then triangle IDB  is right with angle IDB = ADB = 90°

And since  ID = 3  and BI = 5.....then sin (DBI)  = 3/5

And cos DBI = 4/5

But since BE is an angle bisector, then sin (ABC)  = sin (2 * DBI)  =  2 sin(DBI)cos(DBI) = 2 ( 3/5) (4/5) = 24/25

And ADB is also a right triangle with BD  = 4

And since sin ABC = 24/25  then  triangle ADB  is a 7 - 24- 25  Pythagorean right triangle

Then sin ABD = 24/25

So sin of DAB  = 7/25

Using the Law of Sines

AB/ sin 90 = BD / sin (DAB)

AB/ 1  = 4 / (7/25)  =  100/7

And

AD  / sin (ABC)  = BD / sin (DAB)

AD = sin (ABC) * BD / sin (DAB)

AD = (24/25) * 4  / ( 7/25)  = 96/7

And since  angle ADB is right, then so is angle ADC

And since AD is an  angle bisector then  sin (BAD)  = sin (DAC)  = 7/25

So sin (ACD)   = 24/25

So we can find DC using the Law of Sines

DC / sin (DAC)  = AD / sin (ACD)

DC = sin (DAC) * AD / sin (ACD)

DC = (7/25) * ( 96/7) / (24/25)

DC = (7/25) * (96/7) * (25/24)

DC = (96/24)  = 4

So  BC =  BD + DC  = 4 + 4  = 8

And  the area of triangle ABC =  (1/2) (BC) ( AB) * sin ( ABC)  = (1/2) ( 8) ( 100/7) ( 24/25)  = 

(4) ( 4) (24/7)  = 

384/7  units^2

See the figure below to  get a feel  for this :

We can start by factoring 125:

125 = 5^3

Therefore, we can rewrite the equation as follows:

5^3x = a√b

We can also rewrite the right-hand side of the equation as follows:

a√b = a^(1/2)b^(1/4)

Setting the powers of 5 equal on both sides of the equation, we get:

3 = a^(1/2)b^(1/4)

Squaring both sides of the equation, we get:

9 = ab^(1/2)

Therefore, a=9 and b=16.

In conclusion, a=9​ and b=16​.

We can factor the left-hand side of the equation as follows:

(3x - 27)^3 + (27x - 3)^3 = ((3x - 27) + (27x - 3))(((3x - 27)^2 - (3x - 27)(27x - 3) + (27x - 3)^2)

= (30x - 30)((3x - 27)^2 - (3x - 27)(27x - 3) + (27x - 3)^2).

We can factor the right-hand side of the equation as follows:

(3x + 27x - 30)^3 = (60x - 30)^3.

Since (60x−30)3=(30x−30)((60x−30)2), we can write the equation as follows:

(30x - 30)((3x - 27)^2 - (3x - 27)(27x - 3) + (27x - 3)^2) = (60x - 30)((60x - 30)^2).

Since (30x−30)=0, we can cancel it from both sides of the equation, which gives us:

((3x - 27)^2 - (3x - 27)(27x - 3) + (27x - 3)^2) = (60x - 30)^2.

Expanding both sides of the equation, we get:

9x^2 - 162x + 729 - 27x^3 + 196x^2 - 171x + 243 + 729x^2 - 189x + 9 = 3600x^2 - 7200x + 2700

Combining like terms, we get:

2547x^2 - 2316x - 2507 = 0

This quadratic factors as follows:

(x - 1)(2547x + 2507) = 0

The only real solution to this equation is x=1​