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If I cut the wire in half and pass $175$ volts through it, how many milliamps of current will I measure?

\(R=\frac{V}{I}=\frac{500V}{25\cdot 10^{-3}A }\\ R=20 k\Omega\)

The resistance of the two halves of the wire in combination is \(R_2=\frac{20k\Omega}{4}=5k\Omega .\)

\(I=\frac{V}{R_2}=\frac{175V}{5\cdot 10^3 \Omega}\cdot \frac{10^3mA}{A}\\ \color{blue}I=35mA\)

The current of the two halves of the wire in combination is 35mA.

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The product of a set of distinct positive integers is $630$.  

If one of the numbers is 42, what are the other two numbers?  

630 divided by 42 is 15.  The factors of 15 are 3 and 5.    

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I'm not sure if there is an easier way to do this, but one way would involve trigonometric functions.  Since AB = 6, we have an isoceles triangle 5-5-6 at Triangle PAB.  Extend the perpendicular bisector of AB and you have two 3-4-5 right triangles.  Now you can use sin^-1(4/5) to find Angle PAB.  Then you can construct a second right triangle with the hypotenuse being the radius of the circle and one leg being 3, which was the line segment of A to the midpoint of AB.  Then you can use tan^-1 to find the radius.

I am not sure if this is the easiest way, but this was all I could think of.

For polygons with a, b, and c sides to surround a point, 1/a + 1/b + 1/c should equal 1/2.

For example, two regular pentagons and a regular decagon surround a point because 1/5 + 1/5 + 1/10 = 2/10 + 2/10 + 1/10 = 5/10 = 1/2

For an equilateral triangle and a dodecagon,

1/3 + 1/12 + 1/n = 1/2

4/12 + 1/12 + 1/n = 1/2

5/12 + 1/n = 6/12

1/n = 1/12

n = 12

This means that an equilateral triangle and two regular dodecagons completely surround a point.

Since Angle TPO is 33 degrees, then Angle OPU also is 33 degrees, since PO is the angle bisector of Angle TPU.

Therefore, Angle TPU is 66 degrees.

Since tangent lines intercect the circle at 90 degrees, we can form quadrilateral TOUP with Angles T and U being 90 degrees and angle P 66 degrees.

The 4 angles of a quadrilateral add up to 360 degrees, so 360 - 2(90) - 66 = 180 - 66 = 114 degrees.

Central angle TOU is 114 degrees, so the intercepted minor arc TU is 114 degrees.

5x^2 - kx + 8 -2x^2 + 25 = 0

Simplifying,

3x^2 - kx + 33 = 0

a = 3, b = k, c = 33

The formula for the discriminant is b^2 - 4ac.

For the quadratic to have no real solutions, the discriminant must be negative.

So, b^2 - 4ac < 0

Plugging in the values,

k^2 - 4(3)(33) < 0

k^2 -396 < 0

k^2 < 396

For this question, we have to find the largest value of k that is a solution to k^2 < 396.

To maximize k, k should also be positive, (-k)^2 would always have the same solution as k^2

We know that 396 ~ 400, so we can test squares.

20^2 = 400

19 ^ 2 = 361

Therefore, our solution is k = 19.

10 members: 1000 ft^2 wall: 10 min

5 members: 1000 ft^2 wall: 20 min

Half the workforce means double the time.

10 members: 1000 ft^2 wall: 10 min

5 members: 1000 ft^2 wall: 20 min

Half the workforce means double the time.

5x^2 - 2x + 8 - 2x^2 + 6x + 2

Combining Like Terms,

3x^2 + 4x + 10, so

a = 3, b = 4, and c = 10

The formula for the discrimant is b^2 - 4ac

Plugging in the values,

4^2 - 4(3)(10) = 16 - 120 = -104

This means that the quadratic has no real roots.

\(14 = 1\cdot14,2\cdot7\)

\(m = 15,9\)

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Use variables g for grogg and d for dad. We set up the equation d-7 = 9(g-7) and d-4 = 6(g-4).
Solving these equations, we get d = 9g-56 and d=6g-20. Set 9g-56 = 6g-20 and we can get g = 12, plug it back in into the original equation, and groggs dad is 52 years old.

What is the value of the first fraction he wrote?

\(\frac{x}{x+2}\cdot \frac{x+1}{x+3}\cdot\frac{x+2}{x+4}=10\\ \frac{x^2+x}{x^2+7x+12}=10\\x^2+x=10x^2+70x+120\\ 9x^2+69x+120=0\\ 3x^2+23x+40=0\\\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-23 \pm \sqrt{23^2-4\cdot 3\cdot 40} \over 2\cdot 3}\\ \color{blue}x\in\{-2\frac{2}{3},-5\} \)

\(\frac{x}{x+2}\cdot \frac{x+1}{x+3}\cdot\frac{x+2}{x+4}=10\\ x=-5\\ \frac{-5}{-3}\cdot \frac{-4}{-2}\cdot\frac{-3}{-1}=10\\\)

The value of the first fraction he wrote is \(\color{blue}\frac{5}{3}.\)

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\(\gamma =atan \frac{4}{3}\\ \gamma =53.1301^{\circ}\\ \alpha=\arcsin \frac{3}{5}\\ \alpha =33.8699^{\circ}\\ \color{blue}\frac{\alpha}{2}=18.4349^{\circ}\)

\(\angle ADC=(180-53.1301-18.4349)^{\circ}\\ \angle ADC=108.4349^{\circ} \)

\(tan \frac{\\alpha}{2}=\frac {\overline{BD}}{4}\\ \overline{BD}=4\cdot tan\ 18.4349^{\circ}\\ \overline{BD}=1\frac{1}{3}\\ \overline{DC}=3-1\frac{1}{3}\\ \overline{DC}=\frac{5}{3}\)

\(A_{ADC}=\frac{\overline{DC} \cdot \overline{AB}}{2}=\frac{5}{3}\cdot \frac{4}{2}=\frac{20}{6}\\ \color{blue}A_{ADC}=3\frac{1}{3}\)

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The median EC divides the triangle ABD into halves of equal area. If the area of ​​triangle ABC is 14, then the area of ​​triangle BEC is 7.

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Back up !    Edited my answer    x = 3/2     y = 11/12   is the center       r^2 = 32.5 

Here is the graph ( does not look like posted question) :

Using   x,y  and 0,0 

d^2 = (x-0)^2  + (y-0)^2 

Using x,y and -2,1 

d^2 = (x- -2)^2 + (y-1)^2    

Equate these two d^2  s 

    x^2 + y^2  =  x^2 +4x +4     +  y^2 -2y +1   

               0 = 4x -2y + 5    

Now find the distance from x,y to -3,2 

    d^2 = (x- -3)^2  + (y-2)^2  

           =  x^2 + 6x +9  + y^2 -4y + 4         Equate this to the first (or second if you wish) d^2 equation 

        x^2 + y^2  = x^2 + 6x  + 9  + y^2 -4y + 4 

                0 =  6x -4y + 13 

Now solve the system of the two highligted equations 

4x-2y+5 =0 

6x-4y+13 = 0      Multiply the first equation by -2            

-8x +4y-10 = 0       Now add the two equations 

-2x + 3 = 0 

x = 3/2 

then y = 11/2                         (EDITED)          

Re-arrange as

x^2 -6x       +  y^2 +10y    = 18     Complete the square for x and y 

(x-3)^2       +   (y+5)^2     = 18  +9  +25   

(x-3)^2      + (y+5)^2    = 52                <==== this is a circle centered at 3,-5   with radius = sqrt (52) 

Area of a circle = pi r^2 

                         = pi (sqrt (52) )^2   = 52 pi = 163.4 units^2 

The coordinates given the circle are not correkt. (Thanks ElectricPavlov!)

1. Determination of the slope of the chords between the points AO and BA.

\(m_{AO}=\frac{y_A}{x_A}=\frac{1}{-2}=\color{blue}-0.5\\ m_{BA}=\frac{y_B-y_a}{x_B-x_A}=\frac{2-1}{-3-(-2)}=\color{blue}-1\\\)

2. Determination of the coordinates of the centers of the chords BA and AO.

\( P_{BA}(\frac{x_B+x_A}{2},\frac{y_B+y_a}{2})\\ P_{BA}(\frac{-3+(-2)}{2},\frac{2+1}{2})\\ \color{blue}P_{BA}(-2.5,1.5) \)

\(P_{AO}(\frac{x_A+x_O}{2},\frac{y_A+y_O}{2})\\ P_{AO}(\frac{-2+0}{2},\frac{1+0}{2})\\ \color{blue}P_{AO}(-1,0.5)\)

3. Point-direction functions for the center points of the chords BA and AO. The slopes of the function graphs are the negative reciprocals of the slopes of the chords.

\(f_{BA}(x)=-\frac{1}{m_{BA}}(x-x_{PBA})+y_{PBA}\\ f_{BA}(x)=-\frac{1}{-1}(x-(-2.5))+1.5\\ {\color{blue}f_{BA}(x)}=1(x+2.5)+1.5\color{blue}=x+4\\ f_{AO}(x)=-\frac{1}{m_{AO}}(x-x_{P_{AO}})+y_{P_{AO}}\\ f_{AO}(x)=-\frac{1}{-0,5}(x-(-1))+0.5\\ {\color{blue}f_{AO}(x)}=2(x+1)+0.5\color{blue}=2x+2.5 \)

4. Determine the coordinates of point P by equating the two functions.

\(x+4=2x+2.5\\ \color{blue}x_P=1.5\\ y_P=x_P+4=1.5+4\\ \color{blue}y_P=5.5\\ .\\ \color{blue}P(1.5,5.5)\)

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