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We can write the given fraction as follows:

(z + w)/(zw + 1) = (z + w)/((z - w)(1 + zw))

Since the magnitude of z and w are both 1, we know that ∣z−w∣=1. This means that z−w is a complex number with magnitude 1.

We also know that zw is not equal to 1. This means that zw is a complex number with magnitude 1 that is not equal to 1.

Therefore, 1+zw is a complex number with magnitude 1.

Since the denominator of the given fraction is a complex number with magnitude 1, we know that the fraction is a real number.

Alternatively, we can prove that the fraction is a real number by using the following steps:

Let z=a+bi and w=c+di, where a, b, c, and d are real numbers.

Then, zw=(a+bi)(c+di)=(ac−bd)+(ad+bc)i.

Since zw is not equal to 1, we know that ac−bd neq 0.

Therefore, we can divide the numerator and denominator of the given fraction by i, which gives us the following fraction:

(z + w)/(zw + 1) = ((a + bi) + (c + di))/((ac - bd) + (ad + bc)i) = (a + c)/(ac - bd) + (b + d)/(ad + bc)i.

Since ac−bd neq 0, we know that the fraction above is a real number.

We can use the fact that $625=5^4$ and $64=2^6$ to rewrite the given equation as:

(5^4)^x = 2^6

Simplifying the left-hand side using the power of a power rule gives:

5^(4x) = 2^6

We can solve for $x$ by taking the logarithm of both sides with respect to 4, 2 or any other base:

\begin{align*} 4x\log_5(5) &= 6\log_5(2) \ 4x &= \frac{6}{\log_5(5) - \log_5(2)} \ 4x &= \frac{6}{1 - \log_2(5)} \ x &= \frac{3}{2(1-\log_2(5))} \end{align*}

Now, to find $125^x$, we can use the fact that $125=5^3$ and substitute the value we found for $x$:

125^x = (5^3)^(3/(2(1 - log_2(5))) = 5^(9/2(1 - log_2(5)))

Therefore, the value of 125^x is:

5^(9/2(1 - log_2(5))) = 10*sqrt(5)

Hi Tiggsy,

Thanks that was fun.

I won't give away the whole answer but the     e+g = 11132

To find Carissa's average speed during the entire round trip, we need to calculate the total distance she walked and divide it by the total time it took her to walk that distance.

Let's say that the distance from the café to the boba shop is d. Then, the total distance that Carissa walked during the round trip is 2d.

To calculate the total time it took her to walk the round trip, we need to know how much time she spent walking uphill, on flat ground, and downhill. Let's say that she spent t1 hours walking uphill, t2 hours walking on flat ground, and t3 hours walking downhill. Then we have:

t1 = d/100 (since she walks at 100 meters per hour when walking uphill) t2 = d/120 (since she walks at 120 meters per hour when walking on flat ground) t3 = d/150 (since she walks at 150 meters per hour when walking downhill)

The total time it took her to walk the round trip is:

t_total = t1 + t2 + t3 + t3 + t2 + t1 = 2(t1 + t2 + t3) = 2(d/100 + d/120 + d/150)

Therefore 12, Carissa's average speed during the entire round trip is:

average speed = total distance / total time = 2d / (2(d/100 + d/120 + d/150)) = (6000/97) ≈ 61.856 m/hour

Therefore, Carissa's average speed during the entire round trip is approximately 61.856 meters per hour.

idk

The correct answer is 126, but thanks for trying!

Thank you!

We can use the midpoint formula to find the coordinates of point B, which is the midpoint of segment AC.

Let's first find the coordinates of the midpoint M of AC:

M = ((-5 + 4)/2, (2 - 10)/2) = (-0.5, -4)

Now, since BC is twice as long as AB, point B is located one-third of the way from A to M:

B = ((2/3)(-5) + (1/3)(-0.5), (2/3)2 + (1/3)(-4))

Simplifying, we get: B = (-7/2, 0).

https://web2.0calc.com/questions/probability_4869

Try check this

So you could either group them or do complimentary counting, but for this case I think grouping is easier.

Group each child with a adult, so you will have 3 pairs, lets say pairs A,B, and C.

The number of ways to arrange these pairs is 2 ways, however each pair consists of 2 people, so 2^3

=16

(It wasn't specified that the three adults are distinct and the three children are distinct, so I set them as indistinguishable, if they are distinct, just multiply by the arrangements of them.)

Do not post A O P S homework.

1 does not equal 2 though.

The missing operator is TIMES.

1. The answer is 78 degrees.

The two tangents from a point to a circle are congruent, and they form an angle of 180 degrees with the line that passes through the center of the circle and that point.

In the diagram, we know that angle QTA is 47 degrees, and angle RUA is 65 degrees. These two angles are supplementary angles, so they add up to 180 degrees. Therefore, angle QXR is 180 - (47 + 65) = 78 degrees.

2. 

Let $O$ be the circumcenter of $\triangle WXY,$ and let $A$ be the midpoint of $\overline{OX}.$ Then $WA=WX/2=3$ and $OA=OX/2=7/2,$ so by the Pythagorean Theorem, $OW^2=OA^2+WA^2=37/4,$ and $OW=\sqrt{37}/2.$

Let $r$ be the radius of $\odot(WXY).$ Then $XZ=XY-YZ=14-YZ.$ Note that $WZ$ is parallel to the tangent at $X,$ so $\angle WZX=\angle X,$ and $\angle WXA=90^\circ.$ Thus $\triangle WZX\sim\triangle XAO,$ so $WZ/XA=ZX/OA,$ or $WZ=ZX\cdot OA/XA.$ Since $\triangle WXY$ is right-angled at $W,$ we have $WX^2+XY^2=WY^2,$ so $WY=4\sqrt{10},$ and $r=WY/2=2\sqrt{10}.$ Hence $OA^2=r^2+AX^2=40+3^2=49,$ so $OA=7.$ Therefore $WZ=(14-YZ)\cdot 7/3$ and $OZ=OW+WZ=\sqrt{37}/2+7(14-YZ)/3.$

Since $OZ$ is a radius of $\odot(WXY),$ we have $OZ=r,$ which gives 37/2+7(14−��)/3=210.37​/2+7(14−YZ)/3=210​. Solving for $YZ$ gives $YZ = \frac{63}{5}$.

Ok, got it, thank you guest,...stay well

First, let's rewrite the equation by moving all terms to one side:

2x^2 + 22 = x^2 - 12x + 22

Subtract x^2 and 22 from both sides:

x^2 + 12x = 0

Factor the left side:

x(x + 12) = 0

So, the solutions are x = 0 and x = -12, which are r and s. We can now find r^2 + s^2:

r^2 + s^2 = (0)^2 + (-12)^2 = 0 + 144 = 144

To find the probability of each event, we need to know the number of tiles with the desired point value and divide it by the total number of tiles. The distribution of tiles and point values is missing in your message, but I can provide the standard distribution used in the game:

Point value | Tiles

1 point | E ×12, A ×9, I ×9, O ×8, N ×6, R ×6, T ×6, L ×4, S ×4, U ×4

2 points | D ×4, G ×3

3 points | B ×2, C ×2, M ×2, P ×2

4 points | F ×2, H ×2, V ×2, W ×2, Y ×2

5 points | K ×1

8 points | J ×1, X ×1

10 points | Q ×1, Z ×1

a. Drawing a tile that is worth 1 point:

There are 12 + 9 + 9 + 8 + 6 + 6 + 6 + 4 + 4 + 4 = 68 tiles worth 1 point.

The probability of drawing a 1-point tile is 68/100 = 17/25.

b. Drawing a tile that is worth 10 points:

There are 1 + 1 = 2 tiles worth 10 points (Q and Z).

The probability of drawing a 10-point tile is 2/100 = 1/50.

d. Drawing a tile that is worth at least 4 points:

To find the probability of drawing a tile worth at least 4 points, we can count the number of tiles worth 4 points or more. There are 2 + 2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 = 14 tiles worth 4 points or more (F, H, V, W, Y, K, J, X, Q, Z). The probability of drawing a tile worth at least 4 points is 14/100 = 7/50.