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The equation |z - 12| + |z - 5i| = 13 represents the sum of the distances from the complex number z to two fixed points in the complex plane: 12 (on the real number axis) and 5i (on the imaginary number axis).

Here's how to solve for the smallest possible value of ∣z∣ (which represents the distance from the origin):

Triangle Inequality: The sum of the lengths of any two sides of a triangle must be greater than the absolute value of the difference of the remaining side length.

In this case, consider a triangle formed by points z, 12, and 5i. Applying the triangle inequality:

First inequality: |z - 12| + |5i - z| >= |12 - (5i)|

Second inequality: |z - 5i| + |12 - z| >= |5i - 12|

Since both sides of the original equation |z - 12| + |z - 5i| = 13 are positive, we can rewrite it as:

|z - 12| + |z - 5i| = 13

Simplifying the inequalities:

First inequality:

|12 - (5i)| = |-12 + 5i| = sqrt(12^2 + 5^2) = 13 (using the distance formula)

Therefore, |z - 12| + |z - 5i| >= 13 (which is the same as the original equation)

Second inequality:

|5i - 12| = |-12 - 5i| = sqrt((-12)^2 + (-5)^2) = 13 (using the distance formula)

Therefore, |z - 5i| + |12 - z| >= 13

Smallest possible value of |z|:

Since both inequalities we derived involve |z|, the smallest possible value of |z| will occur when both inequalities become equalities.

This means that triangle z - 12 - 5i must be an isosceles triangle with base 12−(−5i)=12+5i and both legs having the same length as ∣z∣.

In other words, point z must be equidistant to both 12 and 5i.

Finding the midpoint:

The midpoint of the line segment connecting 12 and 5i is:

Midpoint = [(12 + 0i) + (0 + 5i)] / 2 = (6 + 2.5i)

Therefore, for the smallest possible value of ∣z∣, point z must coincide with the midpoint, which is (6+2.5i).

Conclusion: The smallest possible value of ∣z∣ is the distance between the origin and the midpoint (6+2.5i), which can be found using the distance formula:

|z| = sqrt(6^2 + (2.5)^2) = sqrt(41)

Therefore, the smallest possible value of z is 6+2.5i​.

Analyzing the game setup, we can see the following:

Landing on a prime number (2 or 3) results in winning that number of dollars.

Landing on a non-prime number (4) results in losing that number of dollars.

We are given the arc size information (120 degrees for prime numbers, 60 degrees for non-prime), but that doesn't directly affect the expected winnings calculation as long as the landing probability on each section is proportional to

its arc size.

Here's how to calculate the expected winnings:

Probability of landing on a prime number:

There are two prime numbers (2 and 3) and their combined arc span is 120 + 120 = 240 degrees.

Since the total degrees is 360 (full circle), the probability of landing on a prime number is (240 degrees / 360 degrees) = 2/3.

Expected value for landing on a prime:

Winning amount for 2 is $2.

Winning amount for 3 is $3.

Expected value when landing on a prime = (2/3) * ($2 + $3) = $4/3

Expected value for landing on a non-prime:

Losing amount for 4 is -$4 (since it's a non-prime).

We can't directly calculate the expected value for non-primes as there's only one number (4) mentioned. There could be other non-prime numbers with different values.

A common approach is to assume an average loss for non-primes. Let's say the average loss for non-primes is also -$4.

Probability of landing on a non-prime:

Probability of non-prime = 1 - (probability of prime) = 1 - (2/3) = 1/3

Overall expected winnings after 1000 games:

Expected winnings = (# of games * (probability of prime * expected value for prime)) + (# of games * (probability of non-prime * expected value for non-prime))

Expected winnings = (1000 games * (2/3 * ($4/3))) + (1000 games * (1/3 * -$4))

Expected winnings = ($1000 * -$4/3) = -$1333.33 (approximately)

You expect to take a loss of 1333.33

First, let's write a quadratic equation to find the two roots. 

We have \(x^{2}+9x-20=0\)

Notice that \(r^2 + s^2 = (r+s)^2-2rs\). 

Now, let's take a break from this. If we had roots r and s, we would have

\((x-r)(x-s)\\ x^2-sx-rx+rs\\ x^2-(r+s)x+rs\)

Using the equation from above, we find that, r+s is equal to -9 and rs is equal to -20. 

Plugging that in to \( (r+s)^2-2rs\), we get\((-9)^2-2(-20) = 81+40 = 121\)

So 121 is our answer!

Thanks! :)\(\)

(x)  + (2x + 10)  + 3(2x +10)  - 10 =  180

x + 2x + 10 + 6x + 20  = 180

9x + 30  =180

9x = 150

x = 150/9  = 50 / 3

Largest angle =     6(50/3) + 20  =  120°

bh =  28

h =  28/b

So.....by the Pythagorean Theorem 

(b/2)^2  + (28/b)^2   =7^2

b^2/4 + 784 / b^2  = 49

b^4 /4 + 784  = 49b^2

b^4/4 - 49/b^2  + 784  = 0

The  only  positive value for b ≈ .25

Only possible perimeter ≈    7 + 7 + .25  ≈  14.25

Thanks for catching my error....corrected  !!!

Each side of the similar triangle is twice  as long as in the orignal triangle

So  perimeter  = ( 16 + 20 + 20)  =  56

I think you made an error... BD^2 = 80, not 16/5. You divided by 5, not multiplied it. But thank you!

https://web2.0calc.com/questions/parallelogram_45

A(-6,2)                             B   (12,2)

D(-6, -2)                          C  (12, -2)

AB  =  12 -  -6 =   18

BC =  2 - - 2 =  4

AB * BC =  18 * 4   =   72   = the area

Triangle BDH  similar to triangle BEC

BD/ DH  = BE  /EC

BD / 5  = BE / EC

BD * EC / 5    = BE

Triangle BEC similar to triangle ADC

BE / EC  = AD / DC

BE / EC  = 16 / DC

BE  / EC = 16/ BD

BE  = 16EC / BD

BD * EC  / 5 = 16 EC / BD

BD ^2  =  80

BD =  4 sqrt 5

[ABC ]  =  BD * AD  =   (4sqrt 5) * 16   =  64  sqrt 5   ≈ 143.1

CORRECTED  !!!

Thanks so much for your effort, even to those who failed!

https://web2.0calc.com/questions/triangle_41241

https://web2.0calc.com/questions/triangles_142

Just a slight mistake by eramsby

The quadratic should be

26x^2 - 354x + 1204 = 0

The solutions  are  x  = 7    and x = 86/13

So y =  5(7) - 28  =  7

And  y = 5(86/13)  - 28  = 66/13

The points are  (7,7)  and (86/13 , 66/13)

Let the center of the circle  = O

Let the midpoint of the large semi-circle = F

Let the midpoint of the  the right semi-circle = D

Let the center of the semi-circle on the left = E

Form triangles EOF   and  DOF

DO =  (5 + r)

FO = (8 - r)

EO = (3 + r)\

DF = 3

EF = 5

By the Law of Cosines  we have two equations

EO^2  = EF^2  + FO^2  - 2(EF * FO) cos (EFO)

DO^2  =  DF^2 + FO^2 - 2(DF * FO) cos ( DFO)

Because  angles EFO  and DFO  are supplementary   cos (DFO) = - cos (EFO)

So re-writing....we have

EO^2  = EF^2  + FO^2  - 2(EF * FO) cos (EFO)

DO^2  =  DF^2 + FO^2 - 2(DF * FO) -cos ( EFO)

EO^2  = EF^2  + FO^2  - 2(EF * FO) cos (EFO)

DO^2  =  DF^2 + FO^2 + 2(DF * FO) cos ( EFO)

(3+r)^2  = 5^2 +(8-r)^2  - 2 (5 * (8-r)) cos (EFO)

(5+ r)^2  = 3^2 + (8-r)^2  + 2( 3 (8-r)) cos EFO

cos EFO  =   [ (3 + r)^2 - 25 - (8-r)^2] /[ -2 * (40 - 5r) ]

cos EFO  = [ (5 + r)^2 - 9 - (8-r)^2] / [ 2 *(24 - 3r) ]

Then

  [ (3 + r)^2 - 25 - (8-r)^2] /[ -2 * (40 - 5r) ] =  [ (5 + r)^2 - 9 - (8-r)^2] / [2 *(24 - 3r) ] 

This is a little sticky to simplify, but WolframAlpha  helps 

We get that

272 / (r-8)  + 49 = 0

272 / (r -8)  = -49

272  = -49 ( r -8)

272 = -49r + 392

-120  =-49r

r =  120 / 49  ≈ 2.449

Since B is on segment AC, we can divide triangle ABC into two right triangles: right triangle ABI and right triangle BCI.

Find the lengths of AI and CI:

We know AB = 6 and BC = 10. Since the semi-circle is constructed on AB, diameter AB is twice the radius of the semi-circle, so the radius of the semi-circle on AB is 6 / 2 = 3. This radius (3) is also the height of right triangle ABI with hypotenuse AB (6) and base AI. Using the Pythagorean Theorem on right triangle ABI:

AI^2 + 3^2 = 6^2 AI^2 = 27 AI = √27 = 3√3

Similarly, the radius of the semi-circle on BC is 10 / 2 = 5. This is the height of right triangle BCI with hypotenuse BC (10) and base CI. Using the Pythagorean Theorem:

CI^2 + 5^2 = 10^2 CI^2 = 75 CI = √75 = 5√3

Radius of the Tangent Circle:

Since the circle is tangent to all three semi-circles, the center of this circle must be at the intersection point of the angle bisectors of triangle ABC. In a right triangle, the angle bisector of the right angle coincides with the median drawn to the hypotenuse, which divides the hypotenuse into two segments with a 1:1 ratio. Therefore, point I, the intersection of AI and CI, is the center of the tangent circle.

We now have a right triangle with legs AI = 3√3 and CI = 5√3, and the hypotenuse is the diameter of the tangent circle (which is twice the radius). Using the Pythagorean Theorem:

(diameter of tangent circle)^2 = (3√3)^2 + (5√3)^2 (diameter of tangent circle)^2 = 48 diameter of tangent circle = √48 = 4√3

Since the diameter is twice the radius, the radius of the tangent circle is:

radius = diameter / 2 = (4√3) / 2 = 2√3

Therefore, the radius of the circle tangent to all three semi-circles is 2√3.

ez

4 hrs at 60 mph = 240 miles

5 hrs at 80 mph = 400 miles

240 + 400 = 640

4 + 5 = 9

his average speed is 640/9 mph(approx. 71.1 mph)

Let's rewrite the equation $n = ab + 3a + 7b$ as $n = a(b+3) + 7b$.

We can factor the equation as $n = (a+7)(b+3) - 21$.

Since $70 \leq n \leq 90$, we have $49 \leq (a+7)(b+3) \leq 77$.

Factoring the bounds, we get $49 = 7 \times 7$ and $77 = 11 \times 7$.

So the possible values for $(a+7)(b+3)$ are $49, 56, 63, 70, 77$.

The factors of these numbers are:

$49$: $(1, 49), (7, 7)$

$56$: $(1, 56), (2, 28), (4, 14), (7, 8)$

$63$: $(1, 63), (3, 21), (7, 9)$

$70$: $(1, 70), (2, 35), (5, 14), (7, 10)$

$77$: $(1, 77), (7, 11)$

Therefore, there are $\boxed{10}$ integers $n$ with $70 \leq n \leq 90$ that can be written as $n = ab + 3a + 7b$ for at least one ordered pair of positive integers $(a, b)$.

Sorry, but I think you missed the fact that B is on AC. Perhaps you could try again? 

Let the point $(x, y)$ be $5$ units away from the point $(2, 7)$. This means that the distance between $(x, y)$ and $(2, 7)$ is $5$ units.

Using the distance formula, we have

\[\sqrt{(x - 2)^2 + (y - 7)^2} = 5.\]

Squaring both sides, we get

\[(x - 2)^2 + (y - 7)^2 = 25.\]

Since the point $(x, y)$ lies on the line $y = 5x - 28$, we substitute $y = 5x - 28$ into the equation above and solve for $x$:

\[(x - 2)^2 + (5x - 35)^2 = 25.\]

\[x^2 - 4x + 4 + 25x^2 - 350x + 1225 = 25.\]

\[26x^2 - 354x + 1205 = 0.\]

The solutions to this quadratic equation for $x$ are $x = 5$ and $x = \frac{23}{13}$. Plugging these values of $x$ back into the equation $y = 5x - 28$, we find that the corresponding $y$ values are $y = 2$ and $y = 1$, respectively.

Therefore, the points that are $5$ units away from $(2, 7)$ and lie on the line $y = 5x - 28$ are $(5, 2)$ and $\left(\frac{23}{13}, 1\right)$.

Let D= (0,0)   Let  B= (40,10)

Slope of line  through DB = ( 10 - 0 ) / (40 -0)  =  1/4

Equation of  this line

y =(1/4)x

Let E = (5,10)  Let F = (8,0)

Slope of line through EF  = (0 -10)/ (8-5)  = -10/3

Equation of this line

y = (-103)(x -8)

y = (-10/3)x + 80/3

To find the x coordinate of P

(1/4)x = (-10/3)x + 80/3

(1/4 + 10/3)X = 80/3

(3/12 + 40/12)x = 80/3

(43/12)x = 80/3

x = (80/3) (12/43)  =  320/43

y coordinate of P  =  (1/4) (320/43)  = 80/43

Height of triangle EBP  =  10 - 80/43 =  350/43

Base of EBP = 40 - 5 = 35

Area of EBP  = (1/2) (35) (350/43)  = 6125/43

[EBP ] / [ ABCD]  =  (6125/43) / [400]  = 245 / 688

why does this look like a problem from AoPS

i hope you aren't trying to cheat on your homework

if you are i'll still give it but at least read the solution and understand it

kay

so

We don't know any other possible outputs from any possible inputs from f(x), so we can only assume that in order to simplify f(x), we have to make the input equal to 3. This gives 5x + 5 = 3, which simplifies to x = -0.4. Then, since f(3) = 4, we have 1/5 * 4 - 5 = 0.8 - 5 = -4.2.

Since the input is -0.4 and the output is -4.2, a point that must be on the graph of f(x) is (-2/5, -21/5).