parmen

From the definition, [ a \mathbin{\spadesuit} (a + 1) = \frac{a^2}{a+1} = 9. ] Multiplying both sides by a+1 yields a2=9(a+1). Substituting a+1 for x and 9 for y into the definition of ♠, we find that this simplifies to 9(a+1)2​=9, which leads to a+1=±3. Therefore, a=−2,2​.

Let f be the probability that a freshman is elected. Then 4f is the probability that a sophomore is elected. Since the sum of the probabilities must be 1, we have f + 4f = 1, which implies f = 1/5. Thus 4f = 4/5. The fraction of the swim team that are sophomores is equal to the probability that a sophomore is elected, which is 4/5. So the answer is 4/5.

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​.