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Let, 

\(\displaystyle a = m^{2}-n^{2},\\\displaystyle b=2mn,\\ \displaystyle c=m^{2}+n^{2},\)

then a, b and c  form a Pythagorean triple.

\(\displaystyle a^{2}+b^{2}=(m^{2}-n^{2})^{2}+(2mn)^{2} \\ \displaystyle =m^{4}+n^{4}-2m^{2}n^{2}+4m^{2}n^{2} \\ \displaystyle =m^{4}+n^{4}+2m^{2}n^{2}=(m^{2}+n^{2})^{2} \\ \displaystyle = c^{2}.\)

What we are looking for then are values of m and n such that m^2 + n^2 = 97.

m^2  = 97 - n^2, so try

n =1 , m^2 = 97 - 1 = 96, m not an integer,

n = 2, m^2 = 97 - 4 = 93  mnai,

n = 3, m^2 = 97 - 9 = 88  mnai,

n = 4, m^2 = 97 - 16 = 81, m = 9, got it.

So, a = 9^2 - 4^2 = 81 - 16 = 65,

b = 2*9*4 = 72.

Check, (for silly mistakes), a^2 + b^2 = 65^2 + 72 ^2 = 4225 + 5184 = 9409 = 97^2. 

It's inversely proportional 
soooo
is 45 members paint a wall in 60 min, then it would take 45x12/45 members to paint 1 wall in 60x 45/12 min.
so 12 members can paint 1 wall in 225 minuutes.

tell me if I'm wrong idk

Sorry, but those are both wrong because my math sire rejected the answers

Problem 3

By the triangle inequality, (5-4 \leq CA \leq 5+4), so 1≤CA≤9. Therefore, there are 9​ possible integer values of CA.

Problem 4

By the Triangle Inequality, we know that 3 <9. We also know that the area of an obtuse triangle is positive, so CA=4 and CA=5. Therefore, the only possible integer values of CA are 3,6,7,8​.

Problem 2

To win a super prize, we can match at least two of the white balls or match the red SuperBall.

Case 1: Matching at least two of the white balls

There are 4 ways to match 3 white balls, 12 ways to match 2 white balls, and 54 ways to match 1 white ball. Therefore, there are 4+12+54=70 ways to match at least two of the white balls.

Case 2: Matching the red SuperBall

There are 8 ways to match the red SuperBall.

Therefore, there are 70+8=78​ ways to win a super prize.

The probability of winning a super prize is 78/(12*11*10*8) = 1/231.

Problem 2

First we will find the slopes of the lines that go through P and Q respectively. The slope of the first line is xP​−xO​yP​−yO​​=4. The slope of the second line is xQ​−xO​yQ​−yO​​=−1. We know the angle between the lines is 180 degrees because they are perpendicular to each other. Then, the slope of the first line is −1 times the reciprocal of the slope of the second line which means the slope is −11​ which is −1. Now we can find the equations of the lines. The equation of the line that goes through P is y−4=−1(x−2) and the equation of the line that goes through Q is y−0=−1(x+8). We can find the coordinates of P by letting y=0 in the first equation. We get 0−4=−1(x−2) and x=6. We can find the coordinates of Q by letting x=0 in the second equation. We get y−0=−1(0+8) and y=−8. The area of triangle OPQ=8*6/2​=24​.

97² = 9409   (c)  

90² = 8100   (a or b)  

17² =   289   (b or a)  

a² + b² must = c²  but, as you can see from the above, it does not.  

The Pythagorean triple with 97 as the hypotenuse is not (90,17,97).  

I recommend always checking an answer, if possible, before posting.  

_.

To find the equation of the perpendicular bisector of the line segment connecting the points (3,2) and (-1,7), we first need to find the midpoint of the line segment and the slope of the line segment.

The midpoint of the line segment is:

((3 - 1)/2, (2 + 7)/2) = (1, 4.5)

The slope of the line segment is:

(7 - 2) / (-1 - 3) = -5/4

The perpendicular bisector of the line segment will have a negative reciprocal slope of -4/5 and will pass through the midpoint of the line segment, (1, 4.5).

Therefore, the equation of the perpendicular bisector of the line segment is:

y - 4.5 = -4/5 (x - 1)

5y - 22.5 = -4x + 4

5y = -4x + 26.5

y = -\frac{4}{5} x + \frac{26.5}{5}

or

y = -0.8x + 5.3

Combining like terms on the right-hand side of the inequality, we get:

2x - 5 <= -9x + 30

Adding 9x to both sides, we get:

11x - 5 <= 30

Adding 5 to both sides, we get:

11x <= 35

Dividing both sides by 11, we get:

x <= 3

Therefore, the solution to the inequality is x <= 3.

Let x be the load that each of the neighbor's small trucks can carry. We know that my large truck can carry a load of at least 500 pounds more than each of her trucks, so my truck can carry at least x+500 pounds.

We also know that my truck can carry no more than 41​ of the total load her four trucks combined can carry, so my truck can carry at most 4x.

Therefore, the greatest load I can be sure that my large truck can carry is 3x+500​ pounds.

a)

There are 8 ways to choose 2 balls out of 8 total.

There are 3 ways to choose 2 black balls out of 3 black.

The probability is 3/8.

b)

The probability of drawing two black balls on the first draw is 3/8​, as you already calculated.

If you don't draw two black balls on the first draw, then you must draw at least one white ball. When you put the white ball back and draw again, you have a 82​ chance of drawing another white ball, and a 83​ chance of drawing a black ball.

Therefore, the probability of drawing two black balls, including the possibility of drawing two black balls on the first draw, is:

(3/8) + (5/8) * (3/8) = 9/16

Let (a,b,97) be the Pythagorean triple we are looking for. We know that a2+b2=972=9409. We also know that a and b must be relatively prime, since the greatest common divisor of the legs of a Pythagorean triple is always 1.

One way to find a primitive Pythagorean triple is to use the Pythagorean formula and experiment with different values for a and b. In this case, we can start by trying to find a value for a that is relatively close to the square root of 9409. We know that 8100​<9409​<10000​, so we can try a=90. Substituting into the Pythagorean formula, we get 902+b2=9409, so b2=9409−902=329. Since 329 is a prime number, we know that b=17. Therefore, the Pythagorean triple with 97 as the hypotenuse is (90,17,97).

Problem 2

We have \begin{align*} f(3!) &= \frac{1}{\log_3 (10!)} \ &= \frac{1}{\log_3 (10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1)} \ &= \frac{1}{\log_3 (3^6\cdot 2^6\cdot 5)} \ &= \frac{1}{6\log_3(3) + 6\log_3(2) + \log_3(5)} \ &= \frac{1}{6+6\cdot 0.699 + 0.431} \ &= \frac{1}{7.130} \ &= \boxed{\frac{100}{713}}. \end{align*}Similarly, \begin{align*} f(5!) &= \frac{1}{\log_5 (10!)} \ &= \frac{1}{\log_5 (5^6\cdot 2^6\cdot 3\cdot 2\cdot 1)} \ &= \frac{1}{6\log_5(5) + 6\log_5(2) + \log_5(3)} \ &= \frac{1}{6+6\cdot 0.431 + 0.176} \ &= \frac{1}{6.607} \ &= \boxed{\frac{150}{991}}. \end{align*}Finally, \begin{align*} f(7!) &= \frac{1}{\log_7 (10!)} \ &= \frac{1}{\log_7 (7^6\cdot 2^6\cdot 5\cdot 3\cdot 2\cdot 1)} \ &= \frac{1}{6\log_7(7) + 6\log_7(2) + \log_7(5)} \ &= \frac{1}{6+6\cdot 0.301 + 0.176} \ &= \frac{1}{6.477} \ &= \boxed{\frac{150}{972}}. \end{align*}

Therefore, f(3!)+f(5!)+f(7!) = 400/2777​.

Problem 1

Case 1: All three numbers are positive. Then ∣a∣=a, ∣b∣=b, and ∣c∣=c, so the expression is equal to [a/a + b/b + c/c + abc/abc = 1 + 1 + 1 + 1 = \boxed{4}.]

Case 2: Two of the numbers are negative and one is positive. Without loss of generality, assume that a and b are negative and c is positive. Then ∣a∣=−a, ∣b∣=−b, and ∣c∣=c, so the expression is equal to [\frac{-a}{-a} + \frac{-b}{-b} + \frac{c}{c} + \frac{-abc}{-abc} = 1 + 1 + 1 - 1 = \boxed{2}.]

Case 3: All three numbers are negative. Then ∣a∣=−a, ∣b∣=−b, and ∣c∣=−c, so the expression is equal to [\frac{-a}{-a} + \frac{-b}{-b} + \frac{-c}{-c} + \frac{-abc}{-abc} = 1 + 1 + 1 + 1 = \boxed{4}.]

Therefore, the possible values of the expression are 2 and 4.

1 wall takes  45 mem * 1 hr  = 45 member hours to paint 

if you have 12 memebers it will take    12 * x = 45 mem hrs 

                                                                x = 45/12 =  3  3/4 hrs 

distance = rate  X time 

d = 40 t 

d = 55(t - 25/60)                    25/60 is the number of hours that is 25 min

equate the two d's

40 t = 55 ( t-25/60) 

-15t = - 22.917 

  t = 1  19/36 hr 

at 40 mi/hr this is       distance = rate  X time 

                                                = 40  X  1 19/36 =  61  1/9 miles 

b  = brother's money     y = your money 

y-5 = b + 5      Given        re-arrange to   y = b + 10 

y+10 = 2 ( b-10)     Given        Sub in y = b+10

(b+10) + 10 = 2b -20

b+20 = 2b -20 

b = 40 dollars                 y = 50 dollars

The integers can be represented as 

x       x+1      x+2       x+3      x+4     x+5    x+6     summed = 7x + 21   and this equals 214 (x+6) 

    7x+21 = 214 (x+6)

    7x + 21 = 214 x + 1284

     - 1263 = 207 x 

          x = 6  7/69                 <===== did you post your question with the correct numbers ?