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In the figure, if MR = MK, the measure of arc MK is 130 degrees, and measure of arc MQ is 23 degrees, then what is angle RPK, in degrees?

Let X be the centre

Mz understanding of the question is that   

and you want 

When two angles stand on the same arc, the angle at the centre is twice the angle on the circumference,

so

None of the other info is relevant.

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For some reason this is not displaying properly, so hee is a picture of it

Let the center of a circle be O

∠AOP = 128º

∠APO = 1/2 (180 - 128) = 26º

∠SPO = 77 - 26 = 51º

arc PS = 180 - 2 * 51 = 78º 

Let f(x) = x^3+bx+c  where b and c are integers. If f(5+\sqrt 2)=0 determine b+c

\((5+\sqrt 2)^3+(5+\sqrt 2)b+c=0\\ \text{If a+b= a constant then I can let one of them be b=0 maybe}\\ If\;\;b=0\;\;then (5+\sqrt 2)^3+c=0\\ c=-(5+\sqrt 2)^3\\ b+c=-(5+\sqrt 2)^3\approx -264 \)

\(However\\ If\;\;c=0\\ (5+\sqrt2)^3+(5+\sqrt2)b=0\\ (5+\sqrt2)b=(5+\sqrt2)^3\\ b=(5+\sqrt2)^2\\ so\\ b+c=(5+\sqrt2)^2\)

The answers are different so b+c does not equal a constant.

I expect you did not copy the question properly.

LaTex:

(5+\sqrt 2)^3+(5+\sqrt 2)b+c=0\\ 
\text{If a+b= a constant then I can let one of them be b=0 maybe}\\
If\;\;b=0\;\;then
(5+\sqrt 2)^3+c=0\\ 
c=-(5+\sqrt 2)^3\\
b+c=-(5+\sqrt 2)^3\approx -264

Thanks Alan, that sure made it easy 

11 definites, 3 uncertain.  only 1 seat available.

P(a person will show)=0.45  

P(a person won't show)=0.55

\(\text{P(no one shows)} \qquad \qquad= \binom{3}{0} * 0.45^0 * 0.55^3\\ \text{P(only one person shows)} = \binom{3}{1} * 0.45^1 * 0.55^2\\ etc\)

You should be able to take it from there.

angle A = 30º          arc DE = 178º          arc BC = 110º

∠30º = 1/2 (arcEC - arcDB) = 1/2 (66 - 6)

The measure of arc CE = 66º

Let's denote a center of a smaller circle with a letter C

∠AOC = ∠CAO = 88 / 2 = 44º 

∠ACO = 180 - 88 = 92º

The measure of arc AOB = 2 * 92 = 184º

omg, hanging my head in shame ... thanks for catching that.  Ron  

_.

arc RK = 360 - 260 = 100º

∠RMK = 100 / 2 = 50º

∠MKQ = 23 / 2 = 11.5º

∠KPM = 180 - (50 + 11.5) = 118.5º

∠RPK = 1/2 (360 - 118.5 * 2) = 61.5º

Thank you!

4 rows of 10 choices for 5 squares     4 * 10 P 5

   plus the middle row                                10 P 4      = (add them together)

Yes, they finish at the same time because the wheels are rotating at the same rpm.

I assumed that they both would finish at the same time but I am unsure.

Missing important info ....please proof read you submission !

Calculate how many three-digit numbers can be formed with the digits 2, 3, 5, 7 and 9, knowing that 5 cannot be the starting number.

First digit   4   choices   (cannot use 5)

second digit  5 choices

third digit     5 choices      

      4 x 5 x 5 = 100 three-digit numbers (assuming numbers can repeat and the number cannot begin with'5')

Yes that is good.

Now in the same amount of time they have both completed 135 degrees.

So that means that in any given period of time they both have covered the same number of degrees.

So who will finish first?

Three consecutive EVEN numbers

x = first one

x+2 = second one

x+4 = third one

x +   x+2    +    x+4 = 270

3x + 6 = 270

3x = 264

x = 264/3 = 88         88   90   and 92

The grid lines in the graph below are one unit apart.  The red parabola shown is the graph of the equation y = ax^2 + bx + c. Find (a,b,c). 

Hello Guest!

Die Gitterlinien in der folgenden Grafik sind eine Einheit voneinander entfernt. Die gezeigte rote Parabel ist der Graph der Gleichung y = ax ^ 2 + bx + c. Finden Sie (a, b, c).

\(P_1(1,-6)\\ P_2(0,-5)\\ P_3(2,-5)\)

\( \color{BrickRed}y = ax^2 + bx + c\\ -6 = a + b + c\\ -5= c\\ -5= 4a + 2b + c\\\)

\(c=-5\)

\( -6 = a + b -5\\ a=-b-1\\ -5= 4a + 2b -5\\ 4a=-2b\\ a=-0.5b\\ -b-1=-0.5b\\ 0.5b=-1\)

\(b=-2\)

\(a=-6-b-c\\ a=-6+2+5 \)

\(a=1\)

\( y = x^2 -2x -5\)

  !

You really can't 'solve' an EXPRESSION such as this....I will assume you omitted part of the EQUATION

8x - 8 = 0     Add 8 to both sides of the equation

8x = 8            divide both sides by 8

x = 1

B)

f(x)  =  √x

Then  f  is not differentiable at  x = 0  because  f'(x)  =  \(\frac12x^{-\frac12}=\frac{1}{2\sqrt{x}}\)   which is undefined when  x = 0

And

f²(x)  =  ( √x )²   =   x

Which is differentiable at  x = 0  because  f²'(x)  =  1  for all values of  x .

Add the equations to get

5x = 15     then x = 3

x+y = 2     then y = -1

I solved for that and got 45pi/4 for Noah and 45pi/2 for Priya. 

Hint:

They have both travelled    135/360   of a whole circumference.

The formula for the circumference of a circle is  2*pi *r

See what you can do that with and get back to us (with what you have done) if you need more help.

A)

Let   f₁(x)  =  x² + 1   and let   f₂(x)  =  kx

In order for  f(x)  to be differentiable at  x = 1, two things must be true at the same time:

First, the graphs must "touch" at  x = 1.  That is,   f₁(1)  =  f₂(1)

Second, the slopes of both graphs at the point where they touch must be identical.  That is,   f₁'(1)  =  f₂'(1)

We can start with either condition and find which value of  k  makes it true, and then check to make sure that same value of  k  makes the other condition true. Let's start with the first condition:

f₁(1)  =  f₂(1)

                            And     f₁(1)  =  1² + 1     and     f₂(1)  =  k(1)

1² + 1  =  k(1)

                            Simplify both sides of the equation.

2  =  k

So we know that  k = 2  satisfies the first condition. Now let's check to make sure it also satisfies the second condition:

f₁'(1)  =  f₂'(1)

                            f₁'(x)  =  2x     so     f₁'(1)  =  2(1)  =  2     and     f₂'(x)  =  k  =  2

2  =  2

The second condition is also satisfied when  k = 2.

So     k = 2

1:1.2:1.5 = 90: 108: 135