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The circumradius of a triangle is the radius of the circle that passes through all three vertices of the triangle. The formula for the circumradius is

R = \frac{abc}{4S},

where a, b, and c are the side lengths of the triangle, and S is the semiperimeter of the triangle, which is

S = \frac{a + b + c}{2}.

In this case, a=5, b=5, and c=6, so S=25+5+6​=8. Therefore, the circumradius is

R = \frac{5 \cdot 5 \cdot 6}{4 \cdot 8} = \boxed{12}.

x^2 + kx - 9x + 25 - 9   =

x² + x(k-9) + 16                    the only two binomial squares that this matches are   ( x-4)²   and  (x+4)²                    

                                                          expanding these :   x^2 - 8x + 16        and    x^2 + 8x + 16 

                                                                          then k - 9 = -8                           k-9 = 8 

                                                                           shows k = 1                            shows k = 17 

Sum the two possible values of k    :    1 + 17 = 18 

Obtuse scalene triangle with:
Sides: a = 3,  b = 5,  c = 7
Using Heron's formula, we calculate the area of the triangle as:
Area: A = 6.495
Perimeter: P = 15
Semi-perimeter: S = 7.5

Inradius: An incircle of a triangle is a tangent circle to each side. We calculate the radius of the incircle as follows:
Radius ==Area / Semi-perimeter
Radius ==6.495 / 7.5 ==0.866

x = worker days needed to complete the job

x/5  =  days for 5 to complete 

x/(5+1) = x/5 - 8     

x/6 = x/5 - 8 

5x/30 - 6x/30 = - 8

     solve for x = 240 worker days 

Now for 28 days less:

240/ ( 5 + n) =  240/5 -28      where n = number of additional workers

240/(5+n) = 20 

n = 7 additional workers needed above the original 5 

x²+6x-5=0

This problem does not require calculus to solve. We cross-multiply and simplify the given equation to get \(yx^2-(y+1)x+y-1=0\). The domain of the function is all real numbers as \(x^2-x+1\) does not have any real solutions, so the discriminant of the equation in \(x\) must always be nonnegative, that is,
$$(y+1)^2-4(y)(y-1) \ge 0 \implies 3y^2-6y-1\le0.$$
The maximum and minimum are simply the roots of the LHS, and their sum is \(-\frac{-6}3=\boxed{-2}\). The values of \(y\) are achievable at the \(x\)-values found in the previous answer.

4(2+2)=16

Really? What makes you think that WAT is from Europe and what makes you think that I am WAT?

In the place of question mark ?, you have to specify the sum total of the 3 dice.

(2/3)R = (4/7) B    →   (7/4)(2/3)R  = B →  (14/12)R = B  →  (7/6)R =  B

And

(2/3)R = (3/5)J   → (5/3)(2/3)R = J →  (10/9)R = J

So

R + (7/6)R + (10/9)R =  1652

(54/54)R + (63/54)R + (60/54)R = 1652

(177/54) R =  1652

R = 1652  ( 54/177)  = $504

B =  (7/6)R  = (7/6)(504)  = $588

J  = (10/9)R = (10/9)(504)  = $560

\(3 \cdot f(x) + 4 \cdot g(x) = h(x) \)

Multiplication of a polynomial by a non-zero constant  doesn't change its  degree

If the degree of h(x) is 9, then the minimum degree of g(x)  must also be 9 .......since the degree of f(x) is less than 9, it doesn't affect the degree of f(x) + g(x)

WAT is from europe

We can solve this problem using complementary counting. We can count the number of ways to roll the dice so that the sum is not 20, and then subtract from the total number of ways to roll the dice.

There are 610 ways to roll the dice in general. If the sum is not 20, then there must be at least one die that rolls a 1. There are 510 ways to roll the dice so that exactly one die rolls a 1, and 410 ways to roll the dice so that exactly two dice roll 1s, and so on.

Note that if exactly k dice roll 1s, then the remaining 10−k dice must all roll 2s or higher. This is because the sum of the numbers on the faces must be at least 20, and if any of the dice rolled a 1, the sum would be less than 20.

Therefore, the number of ways to roll the dice so that the sum is not 20 is equal to: $∑k=110​510−k⋅4k=510(1+4+42+⋯+410)$We can use the formula for a geometric series to evaluate this sum: $1+4+42+⋯+410=4−1411−1​=63,488$Therefore, there are 610−510⋅63,488=492​ ways to roll the dice so the numbers on the faces sum to a total of 20.

The triangle is  isosceles so angles B and C  = 45°

The large  quarter circle  will be tangent  to BC at its midpoint =  (3,3)

The radius of this  circle =  sqrt (3^2 + 3^2)  = sqrt (18)  =  3sqrt (2)

The arc length of this  circle is   (1/4)  *2 * pi * ( 3sqrt (2))  =  (3/2)sqrt (2)  pi            (1)

The radius of each of the smaller arcs at B and C  =  6 - (3)sqrt (2)  

And the combined length of the arcs at B and C =  (1/4) * 2pi * ( 6 - 3sqrt 2)  =  

3pi  - (3/2)sqrt (2)pi      (2)

And the length of the shaded area bordering the hypotenuse  =  sqrt ( 6^2 + 6^2)  - 2 (6 - 3sqrt 2)  = 

6sqrt2 - 12 + 6sqrt 12  =   12 sqrt 2  - 12        (3)

The total perimeter of the  shaded area = (1) + (2) + (3)  = 

(3/2)sqrt( 2) pi   + 3pi - (3/2)sqrt  (2) pi  + 12sqrt (2) - 12  =

3pi + 12(sqrt 2 - 1)    = 14.395 cm ≈  14.4 cm

Since AB and DC are parallel, quadrilateral ABCD is a parallelogram. Thus AC and BD bisect each other at P (in other words, AP=CP and BP=DP). Also, AP+BP=AB and CP+DP=DC, so AP=CP and BP=DP.

We can now pair up congruent triangles:

△ABD≅△CDA (by SAS)

△ADP≅△BCP (by SAS)

Thus there are 2​ pairs of congruent triangles in the diagram.

GA sure does love going hard on 4th graders.

See if  this  helps :  https://web2.0calc.com/questions/quadratic_16582

35   / 2000  =   x /  5000           where x is the  number  of  minutes we are looking for

Cross-multiply

35 * 5000 =  2000 * x            divide both sides by 1000

35 * 5  = 2 * x

175  = 2x

x = 175 /  2    =    87.5  minutes

Oh, you homework cheaters are in the same class?