💡First, calculate the total score Jerry has from the first three tests by multiplying the average by the number of tests: 3 * 94 = 282.Next, add the score Jerry plans to achieve on the fourth test to find the total score for all four tests: 282 + 102 = 384.Then, determine the new average by dividing the total score by the number of tests: 384 / 4 = 96.Finally, find the increase in the average by subtracting the original average from the new average: 96 - 94 = 2.

💡First, I recognize that the car's speed decreases uniformly each second after the brakes are applied. The initial speed is 40 feet per second, and it decreases by 10 feet per second each subsequent second.I can model this situation using an arithmetic sequence where the first term ((a)) is 40 feet and the common difference ((d)) is -10 feet.The sequence of distances traveled each second is as follows:- First second: 40 feet- Second second: (40 - 10 = 30) feet- Third second: (30 - 10 = 20) feet- Fourth second: (20 - 10 = 10) feet- Fifth second: (10 - 10 = 0) feetTo find the total distance traveled until the car stops, I sum these distances:[ 40 + 30 + 20 + 10 + 0 = 100 text{ feet} ]Therefore, the total distance the car travels from the time the brakes are applied until it comes to a complete stop is 100 feet.

💡First, I need to determine how many questions Sasha can complete in one hour, which is given as 15 questions per hour.Next, I calculate the total number of questions she can complete in 2 hours by multiplying her hourly rate by the number of hours she works: 15 questions/hour × 2 hours = 30 questions.Finally, I subtract the number of questions she has completed from the total number of questions she needs to complete to find out how many questions she still needs to finish: 60 questions - 30 questions = 30 questions.

💡Alright, so I have this problem about a prime number ( p = 4k + 1 ). It's divided into three parts, and I need to prove each one step by step. Let me start by understanding what each part is asking for.**Part (a): Prove that there exists an integer ( x ) such that ( x^2 + 1 ) is divisible by ( p ).**Hmm, okay. So I need to show that there's some integer ( x ) where when you square it and add 1, the result is a multiple of ( p ). That means ( x^2 equiv -1 mod p ). I remember something about quadratic residues and maybe Fermat's theorem. Since ( p ) is a prime of the form ( 4k + 1 ), I think this has something to do with it being expressible as a sum of two squares, which might imply that -1 is a quadratic residue modulo ( p ). Let me recall: Euler's criterion says that -1 is a quadratic residue modulo ( p ) if and only if ( p equiv 1 mod 4 ). Since ( p = 4k + 1 ), that must mean -1 is a quadratic residue, so there exists some ( x ) such that ( x^2 equiv -1 mod p ). Therefore, ( x^2 + 1 ) is divisible by ( p ). Okay, that seems straightforward.**Part (b): Prove that it is possible to choose integers ( 0 leq r_1, r_2 < sqrt{p} ) and ( 0 leq s_1, s_2 < sqrt{p} ) such that the numbers ( r_1 x + s_1 ) and ( r_2 x + s_2 ) leave the same remainder when divided by ( p ), and ( (r_1, s_1) neq (r_2, s_2) ).**Alright, so I need to find two different pairs ( (r_1, s_1) ) and ( (r_2, s_2) ) such that ( r_1 x + s_1 equiv r_2 x + s_2 mod p ). This implies that ( (r_1 - r_2)x equiv s_2 - s_1 mod p ). Since ( x ) is invertible modulo ( p ) (because ( p ) is prime and ( x ) is not a multiple of ( p )), this would mean that ( r_1 - r_2 ) is congruent to some multiple of ( (s_2 - s_1) ) times the inverse of ( x ) modulo ( p ).But how do I ensure that such pairs exist? Maybe I can use the pigeonhole principle. The number of possible pairs ( (r, s) ) is ( lfloor sqrt{p} rfloor times lfloor sqrt{p} rfloor ), which is roughly ( p ). Since there are ( p ) possible remainders modulo ( p ), but we have more pairs, by the pigeonhole principle, at least two different pairs must give the same remainder. That makes sense. So, such pairs must exist.**Part (c): Prove that ( p = (r_1 - r_2)^2 + (s_1 - s_2)^2 ).**Okay, from part (b), we have ( (r_1 - r_2)x equiv s_2 - s_1 mod p ). Let me denote ( a = r_1 - r_2 ) and ( b = s_2 - s_1 ). So, ( a x equiv b mod p ). Squaring both sides, we get ( a^2 x^2 equiv b^2 mod p ). From part (a), we know ( x^2 equiv -1 mod p ), so substituting that in, we have ( a^2 (-1) equiv b^2 mod p ), which simplifies to ( -a^2 equiv b^2 mod p ) or ( a^2 + b^2 equiv 0 mod p ).Now, since ( a ) and ( b ) are differences of integers less than ( sqrt{p} ), their absolute values are less than ( sqrt{p} ). Therefore, ( a^2 + b^2 ) is less than ( 2p ). But since ( a^2 + b^2 ) is a multiple of ( p ), the only possibilities are ( a^2 + b^2 = p ) or ( a^2 + b^2 = 0 ). But ( a ) and ( b ) can't both be zero because ( (r_1, s_1) neq (r_2, s_2) ). So, it must be that ( a^2 + b^2 = p ). Therefore, ( p = (r_1 - r_2)^2 + (s_1 - s_2)^2 ).Wait, let me double-check that reasoning. If ( a^2 + b^2 equiv 0 mod p ), then ( p ) divides ( a^2 + b^2 ). Since ( a ) and ( b ) are each less than ( sqrt{p} ), their squares are less than ( p ), so ( a^2 + b^2 ) is less than ( 2p ). Therefore, the only multiple of ( p ) that ( a^2 + b^2 ) can be is ( p ) itself. That makes sense.So, putting it all together:1. For part (a), since ( p equiv 1 mod 4 ), -1 is a quadratic residue modulo ( p ), so there exists an ( x ) such that ( x^2 equiv -1 mod p ).2. For part (b), using the pigeonhole principle, there must be two different pairs ( (r_1, s_1) ) and ( (r_2, s_2) ) such that ( r_1 x + s_1 equiv r_2 x + s_2 mod p ).3. For part (c), manipulating the congruence from part (b) leads us to ( (r_1 - r_2)^2 + (s_1 - s_2)^2 = p ).I think that covers all parts. I should make sure I didn't make any logical jumps without justification, especially in part (c). The key step is recognizing that ( a^2 + b^2 ) must be exactly ( p ) because it's less than ( 2p ) and a multiple of ( p ). That seems solid.Another thing to consider is whether ( r_1, r_2, s_1, s_2 ) are non-negative and less than ( sqrt{p} ). Since we're dealing with remainders, they should be within that range, so the differences ( a ) and ( b ) would naturally be bounded, ensuring that ( a^2 + b^2 ) doesn't exceed ( 2p ).Overall, the reasoning seems consistent and covers all necessary steps. I think I'm ready to present the final proofs.

💡First, determine Jane's age when she stopped baby-sitting. Since she is currently 34 years old and stopped 12 years ago, she was 22 years old when she stopped.Next, calculate the maximum age of a child she could have baby-sat. At the time she stopped, the child's age could not exceed half of Jane's age, which is 22 / 2 = 11 years.Finally, find the current age of the oldest person she could have baby-sat by adding the 12 years that have passed since she stopped. Therefore, the oldest person would now be 11 + 12 = 23 years old.

💡First, I need to understand the pattern of Keun-bae's jump rope routine. He starts with 15 jumps on the first day.On the second day, he jumps twice the number of the first day, which is 2 × 15 = 30 jumps.On the third day, he jumps twice the number of the second day, so 2 × 30 = 60 jumps.Following this pattern, on the fourth day, he will jump twice the number of the third day, which is 2 × 60 = 120 jumps.Therefore, Keun-bae will have to jump rope 120 times on the fourth day.