All Answers 356386 Answers

The length of the hypotenuse of the triangle is 20*sqrt(7).

The length AB is 6 + 8*sqrt(2).

When the same constant is added to the numbers 60, 120, and 160, a three-term geometric sequence arises. What is the common ratio of the resulting sequence?    

In a geometric progression the ratio between succeeding numbers is the same.   

Therefore                                            120 + x         160 + x    

                                                            ––––––   =   ––––––     

                                                             60 + x          120 + x     

Cross multiplying                                 (x + 120)(x + 120)  =  (x + 60)(x + 160)     

                                                             x² + 240x +14,400  =  x² + 220x + 9600       

Subtract x² from both sides                  240x + 14,400  =  220x + 9600    

Subtract 220x from both sides                20x + 14,400  =  9600   

Subtract 10,000 from both sides                            20x  =  – 4800       

                                                                                   x  =  – 240    

Plugging – 240  

into the first ratio                                        – 120   

                                                                 –––––––  =  0.6666 • • •       

                                                                   – 180   

Plugging – 240 

into the second ratio                                   – 80  

                                                                 –––––––  =  0.6666 • • •     

                                                                   – 120              

The common ratio, after     

adding the constant, is                               0.6666 • • •    (or  ²/₃ if you prefer)     

_.    

The answer is BX = 2*sqrt(11).

A terminal zero is a $0$ that appears at the end of a number.    

For example, the number $3,800$ has two terminal zeros.    
How many terminal zeroes does $40 \cdot 6 \cdot 75 \cdot 12$ have?    

          40 • 6 • 75 • 12 = 216,000 ...... so, three terminal zeros    

_.    

The number $100$ has four perfect square divisors, namely $1,$ $4,$ $25,$ and $100.$    
What is the smallest positive integer that has exactly $2$ perfect square divisors?    

That would be four.  Four has exactly two perfect square divisors, namely 1 and 4.   

Note that the problem does not restrict the divisors to only its two perfect squares.   

We ignore 2 as a divisor of 4, same as the problem ignores 10 as a divisor of 100.   

_.    

The answer is BD = 12/5.

The answer is AC = 18.

The answer is CD = 44.

The answer is 45 square units.

Angle CEF is 82 degrees.

Let $x$ and $y$ be complex numbers. If $x + y =2$    

and $xy = 5 - x^2 + xy - y^2$, then what is $x^2 + y^2$?    

ignoring  x + y  =  2    

considering                                 xy  =  5 – x² + xy – y²    

subtract xy from both sides          0  =  5 – x² – y²    

subtract 5 from both sides         – 5  =  – x² – y²    

multiply both sides by – 1              5  =  x² + y²    

                                                      x² + y²  =  5    

_.    

A geometric sequence has 400 terms.    

The first term is 1600 and the common ratio is –1/2.    

How many terms of this sequence are greater than 1000?     

There won't be that many so let's just count them.  

                                                                   1600 • –1/2  =  –800    

                                                                   –800 • –1/2  =    400    

                                                                     400 • –1/2  =  –200    

                                                                   –200 • –1/2  =    100    

                                                                     100 • –1/2  =    –50    

                                                                     –50 • –1/2  =      25    

The numbers are just going to keep dwindling away.     

It's pretty obvious that there is only one term in this      

sequence greater than 1000, i.e., the first one 1600.    

_.   

The answers are A = 1, B = -3, C = 5, D = -2, E = -4, and F = 8.

The sum of the series is 9/4.

The possible values of x are x > 0.

Find the number of bases $b \ge 2$ such that $100_b - 1_b$ is prime.    

The third position to the left is the base raised to the second power position.    

So this problem could be viewed as "For which bases "b" can (b² – 1) be prime?"    

The only base I could find for which this works is 2 ... i.e., 2² – 1 = 3.    

Therefore, the answer to the problem is:  there is only one.    

_.   

The answer is f(17) = -21.

630 divided by 42 = 15     the other two integers, when multiplied = 15    so  either they are   3 and 5   or   1 and 15