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log_a (ab)^2 =

2 log_a (ab) =

2 log_a(a) + 2log_a (b)  =

2 (1) + 2log_a (b) =

2 ( 1 +  log b / log a)

See a similar problem here :  https://web2.0calc.com/questions/geometry_136

https://web2.0calc.com/questions/geometry-question_42617

A line and a circle intersect at A and B as shown below. Find the distance between and A and B.    
The circle is x^2 + y^2 = 1, and the line is y = x.    

x² + y² = 1 draws a circle with radius 1, centered at (0,0).    

y = x draws a straight line through (0,0).    

Since the line goes through the center of the circle, it creates a diameter.    

Since the circle has radius 1, its diameter, the distance between A & B = 2.    

_.    

The area of a trapezoid is $80$. The length of one base is $16$ units greater than the other base, and one base of the trapezoid is $12$. Find the length of the median of the trapezoid.    

M = [ (12) + (12 + 16) ] / 2    

M = 40/2    

M = 20     

The median of a trapezoid is the average of its two bases.    

In this problem, the area of the trapezoid was irrelevant.    

_.    

You can use Guass Formula to find the sum from 1 through n .

(n+1)*n/2=141*140/2=141*70

Since we want to find the prime factors:

141=3*47

70=7*2*5

SO we need the greatest which is \(\left [ 47 \right ]\)

You could do complementary counting and see the ones that are not right.

1,2

1,3

2,1

And since the total ways are 25:

1(Because 25/25=1)-3/25(That are wrong)=22/25

Hope This Helps

Find a six-digit multiple of $64$ that consists only of the digits $2$ and $4$.    

64 • 3,816 = 244,224     

_.    

Let's note that 1/17 is a repeating decimal. 

\(1/17 = 0.\overline{ 0588235294117647}\)

There are 16 repeating digits in the decimal. 

Dividing 4000 by 16, we get

\(4000/16 = 250\)

Since it is divisble, then we can conclude that the last digit of the repeating decimal is the correct answer. 

Thus, 7 is the final answer. 

Thanks! :)

There are four cities: Capital City, Little Village, Mytown, and Yourtown.  The distance from Capital City to Little Village is $5$ miles.  The distance from Capital City to Mytown is $2$ miles.  The distance from Mytown to Yourtown is $3$ miles.  The distance from Little Village to Yourtown is $4$ miles.  Find the distance from Capital City to Yourtown, in miles.    

Little Village ----- 4 ----- Your Town ----- 1 ----- Capital City ----- 2 ----- My Town    

_.    

Let's first focus on the first equation given.

Since we know that \((x+y)^2 = x^2+2xy+y^2\), squaring both sides on the first equation, we get that

\(x^2+2xy+y^2=4\)

Reaaranging the second equation a bit, we find that

\(x^2+xy+y^2=5\)

Wait! These two equations are quite similar to each other. Subtracting the second equation from the first equation, we get that

\(x^2+2xy+y^2=4\\ x^2+1xy+y^2=5 \space \space \space \space -\)

______________________

\(xy=-1\)

Now, let's look at the second equation again. We know that

\(x^2+xy+y^2=5\\ x^2+y^2=5-xy\\ x^2+y^2=5-(-1) = 6\)

So the answer is 6...i think

Thanks! :)

\(Slope = [ b^2 - a^2 ] / [ b - a ] = [ (b -a) (b + a) ] / (b -a) = b + a = 2\)

a + b = 4          square both sides

a^2 + 2ab + b^2 = 16

(a^2 + b^2) + 2ab =16

6 + ab + 2ab = 16

3ab = 10

ab = 10/3

a^2 + b^2 = 6 + ab

a^2 + ^2 - ab = 6

a^3 + b^3  = ( a + b) ( a^2 + b^2 - ab) =  (4)(6)  = 24

One root of the quadratic equation $x^2 + bx -28 = 4x + 14$ is $-7$. What is the value of $b$?    

x² + bx – 28 = 4x + 14    

Change that "bx" to "kx" because I have another use for "b".    

Arrange terms in standard quadratic format ax² + bx + c = 0     

x² + kx – 4x – 42 = 0     

x² + (kx – 4x) – 42 = 0   

x² + (k – 4) x – 42 = 0      and it's given that one root is – 7    

Since one root is – 7, the other root has to be + 6 to get that – 42     

so, factored, we have (x + 7)(x – 6) which multiplied out is x² + x – 42            

& so, (k – 4) x = x which means that k = + 5 or, reverting to the words of the problem, b = 5    

_.    

See a very similar problem  here with Heureka's excellent  approach

https://web2.0calc.com/questions/geometry_136

Call the first term a1   and the common difference, d

We gave these two equations

a1 + 22d = 1/4

a1 + 32d = 1/5      subtract the second equation from the first

-10d = 1/20

d = -1/200

To find the 43rd term :

a43 = a33 + 10d

a43 = a33 + 10 (-1/200)

a43 = 1/5 - 10/200

a43 = 1/5 - 1/20

a43 = 3/20

Only the odd terms can be positive.....so (-8/9)^(n-1)  = (8/9)^(n-1)  when n is odd

We need to solve this

1000 (8/9)^(n -1) > 1        divide both sides by 1000

(8/9)^(n-1) > 1/1000          take the log of both sides

log (8/9)^(n-1) > log (1/1000)    and we  can write

(n-1) log (8/9) > log (1/1000)   rearrange as

n <  log ( 1/1000) /log (8/9) + 1     we divided by a negative - log (8/9) - so, reverse the inequality sign

n <  ≈ 59.6

The 59th term is the last one > 1

So....the number of terms > 1  =   60/2 = 30

 What is the coefficient of x in (x^3 + x^2 + x + 1)(7x^6 + 8x^5 - 4x^4 - 10x^3 + 12x^2 - 2x - 8)?    

Every term in the first expression is multiplied by every term in the second expression.    

Determine the terms that, when multiplied together, result in the "x" coefficient product.   

(x³ + x² + x + 1)(7x⁶ + 8x⁵ – 4x⁴ – 10x³ + 12x² – 2x – 8)    gives you   – 2x    

(x³ + x² + x + 1)(7x⁶ + 8x⁵ – 4x⁴ – 10x³ + 12x² – 2x – 8)    gives you   – 8x    

                              add them together for the final value                      – 10x    

The area of a trapezoid is $80$. The length of one base is $16$ units greater than the other base, and one base of the trapezoid is $12$. Find the length of the median of the trapezoid.    

M = [ (12) + (12 + 16) ] / 2    

M = 40/2    

M = 20    

The median of a trapezoid is the average of its two bases.    

In this problem, the area of the trapezoid is irrelevant.    

_.    

Let P = (x,y)

PA^2 + PB^2 + PC^2  =

(x -2)^2 + ( y- 4)^2  + (x + 3)^2 + (y -1)^2 + (x-1)^2 + ( y-7)^2 ....simplifying, we  have

3x^2 +3y^2 -24y + 80   complete the square on y

3[ (x-0)^2 + y^2  - 24y  + 144] + 80 - 3(144)  =

3 [ ( x-0)^2 + (y- 12)^2 ] - 352  =  3PQ^2 + k

Q= (0 , 12)

k = -352