Boseo

We are given that f(a,b)=2a−3b2+7−5a2+b3 and f(k,−3)=−10. We want to find k.

To find k, we substitute a=k and b=−3 into the expression for f(a,b):

f(k,−3)=2(k)−3(−3)2+7−5(k)2+(−3)3

Simplifying the right side:

f(k,−3)=2k−27+7−5k2+27

Combining like terms:

f(k,−3)=−5k2+2k

We are given that f(k,−3)=−10, so we can set this equal to −10 and solve for k:

−5k2+2k=−10

Move all the terms to one side:

−5k2+2k+10=0

Factor the expression:

$ -(5k^2 - 2k - 10) = 0$

(k+2)(5k−5)=0

Therefore, either k+2=0 or 5k−5=0. Solving for k in each case gives us k=−2 or k=1.

However, we are given that f(k,−3)=−10. Let's evaluate f at each possible value of k:

If k=−2, then f(−2,−3)=2(−2)−3(−3)2+7−5(−2)2+(−3)3=16.

If k=1, then f(1,−3)=2(1)−3(−3)2+7−5(1)2+(−3)3=−10.

Since only f(1,−3)=−10, the solution for k is k=1​.

There are total of 6×5=30 ways of choosing two distinct balls out of 6 balls. Out of these 30 outcomes, only following 10 outcomes result in a product greater than 1

Favorable OutcomesProduct of Numbers on Balls

(1, 2)2

(1, 3)3

(1, 4)4

(1, 5)5

(1, 6)6

(2, 3)6

(2, 4)8

(2, 5)10

(2, 6)12

(3, 4)12

The expected value is the average of the values of each outcome. In this case, this means the average of the products of the numbers on the balls you draw.

So, the expected value is $ \frac{2 + 3 + 4 + 5 + 6 + 6 + 8 + 10 + 12 + 12}{30} = 34/15.

We are given that a1​+a3​=−2 and a2​+a4​=5. In any arithmetic sequence, the common difference between terms is constant. Let d be the common difference.

We can rewrite the first equation as a1​+(a1​+d)=−2 which combines the first and third terms. Simplifying the left side gives 2a1​+d=−2. Similarly, we can rewrite the second equation as a2​+(a2​+d)=5 , which combines the second and fourth terms.

Simplifying the left side gives 2a2​+d=5.

Since each equation represents the sum of consecutive terms in the arithmetic sequence, both equations must have the same common difference, d.

Subtracting the first equation from the second equation gives us (2a2​+d)−(2a1​+d)=5−(−2), which combines the information about d from both equations.

Simplifying both sides gives 2a2​−2a1​=7.

We are solving for a1​, so we want to manipulate this equation to get a1​ by itself. Noticing that both terms on the left-hand side of the equation have a coefficient of 2, we can rewrite it as 2(a2​−a1​)=7. Dividing both sides by 2 gives us a2​−a1​=27​.

Now we can use the fact that the arithmetic sequence has a common difference of d. The difference between a2​ and a1​ is the same as the common difference d. Thus, d=27​.

We can now substitute this value for d in our first equation, 2a1​+d=−2. Substituting 27​ for d gives us 2a1​+27​=−2. Subtracting 27​ from both sides gives us 2a1​=−211​. Dividing both sides by 2 gives us a1​=−411​​.

We can find the sum of all positive integers less than 1000 by itself, and then subtract the sum of the numbers ending in 0, 5, 6, 7, 8, and 9.

The sum of all positive integers less than 1000 is $ \frac{1000 \cdot 999}{2} = 499500$.

The number of integers less than 1000 ending in 0 is 9 (from 10 to 900). The sum of an arithmetic series is the number of terms in the series times the average of the first and last term.

In this case, the sum of the integers ending in 0 is 9⋅(210+900​)=4050. We can do the same for the other digits 5 through 9.

Therefore, the sum of all integers less than 1000 ending in 1, 2, 3, or 4 is 499500−6⋅4050=499500−24300=475200​.

Let x be the constant that is added to each term. Then the three terms become 60+x, 110+x, and 165+x. Note that the common ratio is simply the middle term divided by the first term. Thus, the common ratio is $ \frac{110+x}{60+x}$.

We are given that these terms form a geometric sequence, so the ratio between the second and third terms must equal the ratio between the first and second terms.

Thus, we have the equation 110+x165+x​=60+x110+x​.

Multiplying both sides of the equation by (60+x)(110+x)(165+x) gives us (165+x)(110+x)=(110+x)(60+x).

Expanding both sides gives us 18150+775x+x2=6600+170x+x2. Simplifying the right side gives us 18150+775x+x2=6600+170x+x2.

Subtracting x2 from both sides gives us 18150+775x=6600+170x. Subtracting 170x from both sides gives us 17980+605x=6600.

Subtracting 17980 from both sides gives us 605x=−11380. Dividing both sides by 605 gives us x=−18.8.

Substituting −18.8 for x in our expression for the common ratio, we get a common ratio of $ \frac{110-18.8}{60-18.8} = \frac{91.2}{41.2} = \boxed{\tfrac{45.6}{20.6}}$.

The strategy is to consider how values are scaled when moving from the domain of f to the domain of g.

Since g(x)=f(x/2), taking an input value x in the domain of g and feeding it into f is the same as feeding x/2 into f.

The domain of f is [−8,8], which means the valid inputs for f range from −8 to 8. When we take these values and divide them by 2, we get a range of values from 2−8​=−4 to 28​=4.

Now, these scaled values cover the entire domain of f because f is defined across this range. So the new interval we obtain, [−4,4], has the same width as the original domain [-8, 8]. The width is $ {-8} - {-4} = 4$.

However, the new interval is centered at x=0 instead of x=4. So the resulting domain of g is [−4,4]​.

The interval [-4,4] has width 8.

We can place the diagram in the coordinate plane so that A = (0,s), B = (0,0), C = (s,0), and D = (s,s), where s is the side length of the square.  Let P = (x,y).  Then

x^2 + (y - s)^2 = 1

x^2 + y^2 = 4

(x - s)^2 + y^2 = 9

Solving, we get s = 2 + sqrt(3), x = sqrt(3)/4, y = 1/2 + sqrt(3)/8.  Then by the cosine law, angle APB = 120 degrees.

Since AM is a median, then M is the midpoint of BC. Therefore, BM=MC=6.

Similarly, since BN is a median, then N is the midpoint of AC. Therefore, AN=NC=7.

By the Pythagorean Theorem on right triangle ABM, AB2=AM2+BM2=72+62=85. Then, AB=85​.

Similarly, by the Pythagorean Theorem on right triangle ACN, AC2=AN2+NC2=72+72=98. Then, AC=98​.

Finally, by the Pythagorean Theorem on right triangle ABC, BC^2 = AB^2 + AC^2 = 85 + 98 = 183​.

Therefore, the length of the hypotenuse is BC = sqrt(183).