💡First, I need to understand the initial discount of 40%. This means that after the first discount, the customer pays 60% of the original price.Next, there's an additional 25% discount on the already reduced price. This means the customer now pays 75% of the price after the first discount.To find the final price as a percentage of the original price, I'll multiply the two percentages:0.60 (from the first discount) multiplied by 0.75 (from the second discount) equals 0.45, or 45% of the original price.This means the total discount from the original price is 100% minus 45%, which equals 55%.The store claims a 60% discount, so the difference between the claimed discount and the actual discount is 60% minus 55%, resulting in a 5% difference.

💡First, identify the complements of sets ( A ) and ( B ) with respect to the universal set ( U ).The complement of ( A ) (( C_UA )) consists of elements in ( U ) that are not in ( A ):[C_UA = U setminus A = {4}]Similarly, the complement of ( B ) (( C_UB )) consists of elements in ( U ) that are not in ( B ):[C_UB = U setminus B = {0, 1}]Next, find the union of these two complements:[(C_UA) cup (C_UB) = {4} cup {0, 1} = {0, 1, 4}]Finally, compare the result with the given options to determine the correct answer.

💡First, I need to identify the key points where the lines intersect. The line ( x = 4 ) is a vertical line crossing the x-axis at ( (4, 0) ). The line ( y = 3x ) is a straight line passing through the origin ( (0, 0) ) and rising with a slope of 3.Next, I'll find the point where these two lines intersect. Since ( x = 4 ), I substitute this into the equation ( y = 3x ) to get ( y = 3 times 4 = 12 ). So, the intersection point is ( (4, 12) ).Now, I have three points that form the triangle:1. The origin ( (0, 0) )2. The point where ( x = 4 ) meets the x-axis ( (4, 0) )3. The intersection point ( (4, 12) )To find the area of the triangle, I can use the formula for the area of a triangle with base ( b ) and height ( h ):[text{Area} = frac{1}{2} times b times h]Here, the base ( b ) is the distance from ( (0, 0) ) to ( (4, 0) ), which is 4 units. The height ( h ) is the vertical distance from ( (4, 0) ) to ( (4, 12) ), which is 12 units.Plugging these values into the formula:[text{Area} = frac{1}{2} times 4 times 12 = frac{1}{2} times 48 = 24]So, the area of the triangle is 24.

💡First, I'll start from the end of the sequence of events and work backward to determine how many suckers Jen ate.1. **Taylor to Callie**: Taylor gave the last 5 suckers to Callie. This means Taylor had 5 + 1 = 6 suckers before eating one. 2. **Harmony to Taylor**: Harmony kept 3 suckers and passed the remainder to Taylor. Since Taylor received 6 suckers, Harmony originally had 3 + 6 = 9 suckers.3. **Molly to Harmony**: Molly ate 2 suckers and gave the rest to Harmony. Since Harmony received 9 suckers, Molly had 9 + 2 = 11 suckers before eating 2.4. **Jen to Molly**: Jen ate half of her suckers and gave the rest to Molly. Since Molly received 11 suckers, Jen originally had 11 × 2 = 22 suckers.5. **Suckers Jen ate**: Jen ate half of her suckers, so she ate 22 - 11 = 11 suckers.

💡Alright, so I've got this problem here about boys collecting apples. Each boy either collects 10 apples or 10% of the total apples collected. And there are both types of boys. The question is asking for the minimum number of boys that could have been involved.Okay, let's break this down. First, let's define some variables to make it clearer. Let's say the total number of apples collected is ( k ). Now, there are two types of boys: those who collect 10 apples each and those who collect 10% of the total apples, which would be ( frac{k}{10} ) apples each.Let's denote the number of boys who collect 10 apples as ( n ) and the number of boys who collect ( frac{k}{10} ) apples as ( m ). So, the total number of apples collected by all the boys can be expressed as:[ 10n + m cdot frac{k}{10} = k ]Hmm, okay, let's simplify this equation. Multiply both sides by 10 to eliminate the fraction:[ 100n + mk = 10k ]Now, let's rearrange this to solve for ( k ):[ 100n = 10k - mk ][ 100n = k(10 - m) ][ k = frac{100n}{10 - m} ]So, ( k ) must be an integer because you can't have a fraction of an apple. That means ( frac{100n}{10 - m} ) has to be an integer. Therefore, ( 10 - m ) must be a divisor of 100.Let's list the divisors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100.Now, ( m ) represents the number of boys who collect 10% of the total apples. Since ( m ) is a number of boys, it has to be a positive integer less than 10 (because ( 10 - m ) has to be positive to keep ( k ) positive).Let's test these possible values of ( m ):1. **If ( m = 2 ):** [ 10 - m = 8 ] [ k = frac{100n}{8} = frac{25n}{2} ] For ( k ) to be an integer, ( 25n ) must be even. The smallest ( n ) that satisfies this is 2. [ k = frac{25 times 2}{2} = 25 ] So, total apples ( k = 25 ). Then, the number of boys who collect 10% is ( m = 2 ), and each of them collects ( frac{25}{10} = 2.5 ) apples. Wait, that's not possible because you can't have half an apple. So, ( m = 2 ) doesn't work.2. **If ( m = 4 ):** [ 10 - m = 6 ] [ k = frac{100n}{6} = frac{50n}{3} ] For ( k ) to be an integer, ( 50n ) must be divisible by 3. The smallest ( n ) that satisfies this is 3. [ k = frac{50 times 3}{3} = 50 ] So, total apples ( k = 50 ). Then, each boy collecting 10% would get ( frac{50}{10} = 5 ) apples. That works because 5 is an integer. So, with ( m = 4 ) and ( n = 3 ), we have a valid solution. Total number of boys is ( m + n = 4 + 3 = 7 ).3. **If ( m = 5 ):** [ 10 - m = 5 ] [ k = frac{100n}{5} = 20n ] For ( k ) to be an integer, ( n ) can be any integer. Let's take the smallest ( n = 1 ). [ k = 20 times 1 = 20 ] Each boy collecting 10% would get ( frac{20}{10} = 2 ) apples. That works. So, with ( m = 5 ) and ( n = 1 ), we have another valid solution. Total number of boys is ( m + n = 5 + 1 = 6 ).4. **If ( m = 1 ):** [ 10 - m = 9 ] [ k = frac{100n}{9} ] For ( k ) to be an integer, ( 100n ) must be divisible by 9. The smallest ( n ) that satisfies this is 9. [ k = frac{100 times 9}{9} = 100 ] Each boy collecting 10% would get ( frac{100}{10} = 10 ) apples. That works. So, with ( m = 1 ) and ( n = 9 ), we have another valid solution. Total number of boys is ( m + n = 1 + 9 = 10 ).From the above, the smallest number of boys is when ( m = 5 ) and ( n = 1 ), giving a total of 6 boys.But wait, let's double-check this. If there are 5 boys collecting 10% and 1 boy collecting 10 apples, the total apples would be:[ 5 times frac{k}{10} + 1 times 10 = k ][ frac{5k}{10} + 10 = k ][ frac{k}{2} + 10 = k ][ 10 = k - frac{k}{2} ][ 10 = frac{k}{2} ][ k = 20 ]So, each of the 5 boys collects ( frac{20}{10} = 2 ) apples, and the 1 boy collects 10 apples. That adds up to 20 apples in total, which works.Is there a way to get fewer than 6 boys? Let's see.If ( m = 6 ):[ 10 - m = 4 ][ k = frac{100n}{4} = 25n ]For ( k ) to be an integer, ( n ) can be any integer. Let's take ( n = 1 ):[ k = 25 ]Each boy collecting 10% would get ( frac{25}{10} = 2.5 ) apples, which isn't possible. So, ( m = 6 ) doesn't work.If ( m = 3 ):[ 10 - m = 7 ][ k = frac{100n}{7} ]For ( k ) to be an integer, ( 100n ) must be divisible by 7. The smallest ( n ) that satisfies this is 7.[ k = frac{100 times 7}{7} = 100 ]Each boy collecting 10% would get ( frac{100}{10} = 10 ) apples. So, with ( m = 3 ) and ( n = 7 ), total boys would be 10, which is more than 6.If ( m = 0 ):But the problem states there are both types of boys, so ( m ) can't be 0.If ( m = 10 ):[ 10 - m = 0 ]Which would make ( k ) undefined, so that's not possible.So, after checking all possible values, the minimum number of boys is 6.

💡Okay, so I have this function f(x) = lg|x|, and I need to figure out the range of real numbers a such that f(1) < f(a). Hmm, let me start by understanding what this function does.First, lg|x| is the logarithm base 10 of the absolute value of x. So, it's defined for all real numbers except x = 0 because the logarithm of zero is undefined. Also, since it's the absolute value, the function is even, meaning f(-x) = f(x). That's useful because it tells me the function behaves the same way for positive and negative x values.Now, let's compute f(1). Since f(x) = lg|x|, f(1) would be lg|1|, which is lg(1). I remember that log base 10 of 1 is 0 because 10^0 = 1. So, f(1) = 0.The problem states that f(1) < f(a), which translates to 0 < f(a). So, I need to find all real numbers a such that f(a) is greater than 0.Given that f(a) = lg|a|, the inequality becomes lg|a| > 0. I need to solve this inequality for a.I know that lg|a| > 0 means that |a| > 10^0, which is |a| > 1. Because logarithm base 10 of a number is greater than 0 when the number is greater than 1. So, |a| > 1.Now, solving |a| > 1 gives me two cases: either a > 1 or a < -1. That's because the absolute value of a is greater than 1 if a is either greater than 1 or less than -1.Let me double-check to make sure I haven't missed anything. The function f(x) = lg|x| is defined for all x except 0. It's symmetric about the y-axis because it's an even function. The logarithm function increases as its argument increases, so as |x| increases, f(x) increases.Therefore, f(a) will be greater than f(1) when |a| is greater than |1|, which is 1. So, |a| > 1, leading to a > 1 or a < -1.I think that's correct. I don't see any mistakes in my reasoning. So, the range of a is all real numbers greater than 1 and less than -1.**Final Answer**The range of the real number ( a ) is boxed{a > 1 text{ or } a < -1}.