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1st attempt:   The triangle ABC has a 90° angle at a vertex A.   

The ratio of the area of a triangle QBS to the area of a triangle RQC is   5/9    or  5 : 9   ( 0.555555 )

2nd attempt:   The triangle ABC has a 45° angle at a vertex A.

The result is the same:   the ratio is 5 / 9

The ratio of the area of triangle QBC to the area of triangle RQC is 4/7.

a = -3, b = 3, c = 5, so a + 2b + 3c = (-3) + 2*(3) + 3(5) = 18.

Since are looking for a cubic polynomial, let it be ax^3 + bx^2 + cx + d = 0.  So

\(a(\sqrt[3]{2} + \sqrt[3]{4})^3 + b(\sqrt[3]{2} + \sqrt[3]{4})^2 + c(\sqrt[3]{2} + \sqrt[3]{4}) + d = 0\)

This expands to

\(a (2 + 3 \sqrt[3]{2^2} \sqrt[3]{4} + 3 \sqrt[3]{2} \sqrt[3]{4^2} + 4) + b(\sqrt[3]{2^2} + 2 \sqrt[3]{2} \sqrt[3]{4} + \sqrt[3]{4^2}) + c(\sqrt[3]{2} + \sqrt[3]{4}) + d = 0\)

Expanding everything out, and comparing the coefficients, we get

a + 3b + 3c + d = 16,
-6b + 3c + 3d = -24,
c - 3d = 22,
d = -6.

The solution to this system is a = 1, b = 3, c = 4, d = -6, so the cubic is x^3 + 3x^2 + 4x - 6.

Letting x go to 2, we find that A + 2B = 4.  Letting x go to -1, we find 2A - B = 3.  Therefore, A = 2 and B = -1.

Let the width shown = w     and  the length shown be  h   (for height)

\(\frac{2w}{h}=\frac{h}{w}\\ 2w^2=h^2\\ \sqrt{2}w=h\\ \text{so the ratio of width to length as shown is}\qquad 1:\sqrt2\)

I misunderstood the question.  

The diameter of the semi circle is:   D = 2{ sin60° * sqrt(4) }   

Use the quadratic formula to find the roots

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)       a = 1   b = -6   c = 2    

x = 3 +- sqrt 7          real 

Test the quadratic equation:

 \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a=1

b=-6

c=2

solve the quadratic equation and see if they're real or not :) the roots aren't real if they include the square root of a negative nuumber.

The prime factorization of 12! is 2^10(3^5)(5^2)(7)(11). Then, take the primes that are multiplied more than once, so 2^10, 3^4(leave the last 3 behind i will explain later), and 5^2. We want to make each exponent even because it must be a perfect square, which is why i didn't include the 5th 3. Therefore, the largest perfect square that divides 12! is 2^10 * 3^4 *5^2 = 2073600.

sqrt(x+3)-1=x

SQRT(X+3) = X+1     SQUARE BOTH SIDES

x+3 = x^2 + 2x + 1    gather like terms

x^2 + x -2 = 0

(x+2)(x-1) = 0     so x = -2 or +1   -2 does not work -throw out   

The answer is 15, which I got.

that is incorrect.

Anyone?? :(

Angle ACP is 25 degrees.

The last sentence is messed up, but it's supposed to be if \(\angle APC = 115^\circ\), then what is \(\angle ACP\) in degrees?

For the first problem, of course, |zw|^2 = |z|^2 |w|^2 = 64 and |z/w|^2 = |z|^2/|w|^2 = 4.

Then we can use the identities
|z + w|^2 = (z + wi)(z - wi) = |z|^2 + |z||w| + |w|^2 = 4^2 + 4*2 + 2^2 = 28
and
|z - w|^2 = (z - wi)(z + wi) = |z|^2 - |z||w| + |w|^2 = 4^2 - 4*2 + 2^2 = 12

For the second problem, |zw|^2 = |z|^2 |w|^2 = 100 and |z/w|^2 = |z|^2/|w|^2 = 25/4.

Then we can use the identities
|z + w|^2 = (z + wi)(z - wi) = |z|^2 + z bar(w) + bar(z) w + |z||w| + |w|^2 = 25 + (6 + 8i) + (6 - 8i) + 5*2 + 2^2 = 51
and
|z + w|^2 = (z + wi)(z - wi) = |z|^2 - z bar(w) - bar(z) w - |z||w| + |w|^2 = 25 - (6 + 8i) - (6 - 8i) - 5*2 + 2^2 = 7

Let w = a + bi and z = c + di.  Then

\(\dfrac{w + z}{1 + wz} = \dfrac{a + c + bi + di}{1 + (a + bi)(c + di)}\)

To express this in rectangular form, we can multiply the numerator and denominator by the conjugate:

\(\dfrac{a + c + bi + di}{1 + (a + bi)(c + di)} = \dfrac{(a + c + bi + di)((1 - (a + bi)(c + di))}{(1 + (a + bi)(c + di))(1 - (a + bi)(c + di))}\)

The denominator simplifies to (1 - (a^2 + b^2)(c^2 + d^2)), which is real.  The numerator simplifies to a^2 - b^2 + c^2 - d^2, which is also real.  Therefore, the complex number (w + z)/(1 + wz) is real.

Geometric mean   Multiply   'n' numbers together then take the 'n^{th '}  root of the result....

Solve for x:
log(log(log(x)/log(2))/log(3))/log(4) = 0

Multiply both sides by log(4):
log(log(log(x)/log(2))/log(3)) = 0

Cancel logarithms by taking exp of both sides:
log(log(x)/log(2))/log(3) = 1

Multiply both sides by log(3):
log(log(x)/log(2)) = log(3)

Cancel logarithms by taking exp of both sides:
log(x)/log(2) = 3

Multiply both sides by log(2):
log(x) = 3 log(2)

3 log(2) = log(2^3) = log(8):
log(x) = log(8)

Cancel logarithms by taking exp of both sides:
 
x = 8

See previous post....

https://web2.0calc.com/questions/please-help-me-graph-the-function-on-the-coordinate

Pick some x values from,   say -5 to 5

    calculate the f(x) values and plot them, then 'sketch' the graph by connecting the dots....

Here is what it looks like:

https://www.desmos.com/calculator/b79a0cvuku

I'm sorry. I copied the question wrong...