💡Okay, so I have this function f(x) = 2^{2x} - 2^{x+1} + 3. I need to find the maximum value of f(x) when x is in the interval [-1, 2], and then also find the maximum and minimum values of f(x) on the interval [m, 0]. Hmm, let me try to figure this out step by step.First, looking at the function f(x) = 2^{2x} - 2^{x+1} + 3. It seems a bit complicated with the exponents, but maybe I can simplify it by substitution. Let me set t = 2^x. Then, since 2^{2x} is (2^x)^2, that would be t^2. Similarly, 2^{x+1} is 2*2^x, which is 2t. So substituting these into the function, I get f(x) = t^2 - 2t + 3. That looks simpler.So now, f(x) can be written as h(t) = t^2 - 2t + 3. Hmm, this is a quadratic function in terms of t. Quadratic functions are easier to analyze because I can find their vertex and determine if it's a maximum or minimum. The general form of a quadratic is at^2 + bt + c, so here a = 1, b = -2, c = 3.The vertex of a quadratic occurs at t = -b/(2a). Plugging in the values, t = -(-2)/(2*1) = 2/2 = 1. So the vertex is at t = 1. Since the coefficient of t^2 is positive (1), the parabola opens upwards, meaning the vertex is a minimum point. Therefore, the minimum value of h(t) is at t = 1, and it's h(1) = (1)^2 - 2*(1) + 3 = 1 - 2 + 3 = 2.Alright, so the minimum value of h(t) is 2. But the question is about the maximum value of f(x) on the interval [-1, 2]. Since h(t) is a parabola opening upwards, its maximum on a closed interval will occur at one of the endpoints. So I need to find the corresponding t values for x = -1 and x = 2.When x = -1, t = 2^{-1} = 1/2. When x = 2, t = 2^2 = 4. So the interval for t is [1/2, 4]. Now, I need to evaluate h(t) at both endpoints and see which one is larger.Calculating h(1/2): (1/2)^2 - 2*(1/2) + 3 = 1/4 - 1 + 3 = 1/4 + 2 = 9/4 = 2.25.Calculating h(4): (4)^2 - 2*(4) + 3 = 16 - 8 + 3 = 11.Comparing h(1/2) = 2.25 and h(4) = 11, clearly 11 is larger. So the maximum value of f(x) on the interval [-1, 2] is 11.Okay, that seems straightforward. Now, moving on to the second part: finding the maximum and minimum values of f(x) on [m, 0]. So x is between m and 0. I need to figure out how this interval affects the function.Again, using the substitution t = 2^x. Since x is between m and 0, and m is presumably less than or equal to 0, because the interval is [m, 0]. So, 2^x is an increasing function, which means as x increases, t increases. So when x = m, t = 2^m, and when x = 0, t = 2^0 = 1. Therefore, the interval for t is [2^m, 1].Now, h(t) is still the quadratic function t^2 - 2t + 3, which we know has its minimum at t = 1. Since the interval for t is [2^m, 1], and the function is a parabola opening upwards, the minimum value on this interval will still be at t = 1, which is 2.But what about the maximum? Since the parabola opens upwards, the maximum on the interval [2^m, 1] will occur at the left endpoint, which is t = 2^m. So the maximum value of h(t) on this interval is h(2^m) = (2^m)^2 - 2*(2^m) + 3 = 2^{2m} - 2^{m+1} + 3.Wait, let me verify that. If t is in [2^m, 1], and since the parabola opens upwards, the function decreases from t = 2^m to t = 1. So yes, the maximum is at t = 2^m and the minimum is at t = 1.But hold on, is 2^m always less than 1? Since m is less than or equal to 0, 2^m is 1/(2^{|m|}), which is between 0 and 1. So yes, 2^m is less than 1, so the interval [2^m, 1] is valid.Therefore, on the interval [m, 0], the function f(x) has a minimum value of 2 and a maximum value of 2^{2m} - 2^{m+1} + 3.Let me just recap to make sure I didn't miss anything. For part (1), I transformed the function into a quadratic in terms of t, found the interval for t based on x in [-1, 2], evaluated the quadratic at the endpoints, and found the maximum value at t = 4, which corresponds to x = 2.For part (2), I considered the interval [m, 0], transformed it into the corresponding t interval [2^m, 1], recognized that the quadratic has its minimum at t = 1, and the maximum at t = 2^m. So the maximum is h(2^m) and the minimum is h(1) = 2.I think that's all. I don't see any mistakes in my reasoning. The key was substituting t = 2^x to turn the function into a quadratic, which is easier to analyze for maxima and minima.**Final Answer**- The maximum value of ( f(x) ) when ( x in [-1, 2] ) is (boxed{11}).- The minimum value of ( f(x) ) on ([m, 0]) is (boxed{2}), and the maximum value is (boxed{2^{2m} - 2^{m+1} + 3}).

💡Okay, so I have this problem where I need to find the range of values for ( a ) such that the line ( x - y - a = 0 ) intersects the circle ( C: (x - 1)^2 + y^2 = 2 ) at a common point. Hmm, let me think about how to approach this.First, I remember that for a line and a circle to intersect, the distance from the center of the circle to the line must be less than or equal to the radius of the circle. That makes sense because if the distance is greater than the radius, the line won't touch the circle at all. So, I need to find the distance from the center of the circle to the given line and set up an inequality based on the radius.Let me write down the equation of the circle again: ( (x - 1)^2 + y^2 = 2 ). From this, I can see that the center of the circle is at ( (1, 0) ) and the radius is ( sqrt{2} ) because the standard form of a circle is ( (x - h)^2 + (y - k)^2 = r^2 ).Now, the line is given by ( x - y - a = 0 ). I need to find the distance from the center ( (1, 0) ) to this line. The formula for the distance ( d ) from a point ( (x_0, y_0) ) to the line ( Ax + By + C = 0 ) is:[d = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}]In this case, the line is ( x - y - a = 0 ), so ( A = 1 ), ( B = -1 ), and ( C = -a ). Plugging in the center ( (1, 0) ):[d = frac{|1 cdot 1 + (-1) cdot 0 - a|}{sqrt{1^2 + (-1)^2}} = frac{|1 - a|}{sqrt{2}}]Okay, so the distance from the center to the line is ( frac{|1 - a|}{sqrt{2}} ). Now, for the line to intersect the circle, this distance must be less than or equal to the radius ( sqrt{2} ). So, I set up the inequality:[frac{|1 - a|}{sqrt{2}} leq sqrt{2}]To simplify this, I can multiply both sides by ( sqrt{2} ):[|1 - a| leq 2]This absolute value inequality means that ( 1 - a ) is between ( -2 ) and ( 2 ). So, I can write:[-2 leq 1 - a leq 2]Now, I need to solve for ( a ). Let's subtract 1 from all parts of the inequality:[-2 - 1 leq -a leq 2 - 1][-3 leq -a leq 1]Next, I'll multiply all parts by ( -1 ). But I have to remember that multiplying an inequality by a negative number reverses the inequality signs:[3 geq a geq -1]Which can also be written as:[-1 leq a leq 3]So, the range of values for ( a ) is from ( -1 ) to ( 3 ). Let me just double-check my steps to make sure I didn't make any mistakes.1. I identified the center and radius of the circle correctly.2. I used the distance formula from a point to a line accurately.3. I set up the inequality correctly based on the radius.4. I solved the inequality step by step, remembering to reverse the inequality signs when multiplying by a negative.Everything seems to check out. I think I'm confident with this answer.**Final Answer**The range of values for ( a ) is boxed{[-1, 3]}.

💡First, I need to determine how many days Larry jogs in the first week. I know that Larry jogs for 30 minutes each day and that he jogs for a total of 4 hours over two weeks. I also know that he jogs for 5 days in the second week.To make the calculations easier, I'll convert the total jogging time from hours to minutes. Since there are 60 minutes in an hour, 4 hours is equal to 4 * 60 = 240 minutes.Next, I'll calculate the total time Larry spends jogging in the second week. He jogs for 5 days, and each day he jogs for 30 minutes. So, the total time for the second week is 5 * 30 = 150 minutes.Now, I'll subtract the time spent jogging in the second week from the total time to find out how much time he spends jogging in the first week:240 minutes (total) - 150 minutes (second week) = 90 minutes (first week).Finally, to find out how many days Larry jogs in the first week, I'll divide the total time he spends jogging in the first week by the time he jogs each day:90 minutes (first week) / 30 minutes/day = 3 days.Therefore, Larry jogs for 3 days in the first week.

💡First, I'll define the amount Lily spent on the shirt as ( S ).Lily started with 55 in her account. After purchasing the shirt, she had:[55 - S]She then spent three times the amount she spent on the shirt at another shop, which is:[3S]After this second purchase, the remaining amount in her account is:[(55 - S) - 3S = 55 - 4S]According to the problem, Lily has 27 left in her account. Therefore, I can set up the equation:[55 - 4S = 27]To solve for ( S ), I'll first subtract 55 from both sides:[-4S = 27 - 55][-4S = -28]Next, I'll divide both sides by -4:[S = frac{-28}{-4}][S = 7]Thus, Lily spent 7 on the shirt.

💡To find the sum of the six smallest distinct positive integer multiples of 11, I start by identifying these multiples. They are 11, 22, 33, 44, 55, and 66.Next, I factor out the common factor of 11 from each term, which simplifies the expression to 11 multiplied by the sum of the first six positive integers: 1, 2, 3, 4, 5, and 6.Using the formula for the sum of the first n positive integers, which is ( frac{n(n+1)}{2} ), I substitute n with 6 to calculate the sum: ( frac{6 times 7}{2} = 21 ).Finally, I multiply this sum by 11 to find the total sum of the multiples: ( 11 times 21 = 231 ).

💡Okay, so I have this problem here where I need to figure out the relationship between a line and a circle. The line is given by the equation ( mx + y - m - 1 = 0 ) where ( m ) is a real number, and the circle is ( x^2 + y^2 = 2 ). The options are whether they are disjoint, tangent, intersect or tangent, or just intersect. Hmm, let me think about how to approach this.First, I remember that to find the relationship between a line and a circle, we can use the concept of distance from the center of the circle to the line. If the distance is greater than the radius, they are disjoint; if it's equal, they are tangent; and if it's less, they intersect at two points.So, the circle ( x^2 + y^2 = 2 ) has its center at (0,0) and radius ( sqrt{2} ). The line is ( mx + y - m - 1 = 0 ). I need to find the distance from the center (0,0) to this line.The formula for the distance ( d ) from a point ( (x_0, y_0) ) to the line ( ax + by + c = 0 ) is:[d = frac{|ax_0 + by_0 + c|}{sqrt{a^2 + b^2}}]In our case, the line is ( mx + y - m - 1 = 0 ), so ( a = m ), ( b = 1 ), and ( c = -m - 1 ). The point is (0,0), so plugging into the formula:[d = frac{|m cdot 0 + 1 cdot 0 - m - 1|}{sqrt{m^2 + 1^2}} = frac{| -m - 1 |}{sqrt{m^2 + 1}} = frac{|m + 1|}{sqrt{m^2 + 1}}]Okay, so the distance is ( frac{|m + 1|}{sqrt{m^2 + 1}} ). Now, we need to compare this distance with the radius ( sqrt{2} ).So, let's set up the inequality:[frac{|m + 1|}{sqrt{m^2 + 1}} leq sqrt{2}]Because if the distance is less than or equal to the radius, the line either intersects the circle or is tangent to it. If it's greater, they are disjoint.Let me square both sides to eliminate the square roots:[left( frac{|m + 1|}{sqrt{m^2 + 1}} right)^2 leq (sqrt{2})^2][frac{(m + 1)^2}{m^2 + 1} leq 2]Multiply both sides by ( m^2 + 1 ) (which is always positive, so the inequality sign doesn't change):[(m + 1)^2 leq 2(m^2 + 1)]Expand the left side:[m^2 + 2m + 1 leq 2m^2 + 2]Bring all terms to one side:[m^2 + 2m + 1 - 2m^2 - 2 leq 0][- m^2 + 2m - 1 leq 0][- (m^2 - 2m + 1) leq 0][- (m - 1)^2 leq 0]Hmm, ( (m - 1)^2 ) is always non-negative, so multiplying by -1 makes it non-positive. Therefore, ( - (m - 1)^2 leq 0 ) is always true for all real ( m ). That means the inequality ( frac{|m + 1|}{sqrt{m^2 + 1}} leq sqrt{2} ) holds for all real ( m ).Wait, so that would mean the distance is always less than or equal to the radius, so the line either intersects the circle or is tangent to it. So the relationship is either intersect or tangent.But let me double-check. Maybe I made a mistake in the algebra.Starting from:[(m + 1)^2 leq 2(m^2 + 1)][m^2 + 2m + 1 leq 2m^2 + 2][0 leq m^2 - 2m + 1][0 leq (m - 1)^2]Ah, I see. When I moved everything to the other side, I should have:[m^2 + 2m + 1 - 2m^2 - 2 leq 0][- m^2 + 2m - 1 leq 0][m^2 - 2m + 1 geq 0][(m - 1)^2 geq 0]Which is always true because a square is always non-negative. So, the inequality ( frac{|m + 1|}{sqrt{m^2 + 1}} leq sqrt{2} ) holds for all real ( m ), meaning the distance is always less than or equal to the radius.Therefore, the line either intersects the circle at two points or is tangent to it. So, the correct answer should be option C: Intersect or Tangent.But wait, let me think again. Is there a specific value of ( m ) where the line is tangent to the circle? Let's see.If the distance equals the radius, then it's tangent. So, when does ( frac{|m + 1|}{sqrt{m^2 + 1}} = sqrt{2} )?Squaring both sides:[frac{(m + 1)^2}{m^2 + 1} = 2][(m + 1)^2 = 2(m^2 + 1)][m^2 + 2m + 1 = 2m^2 + 2][0 = m^2 - 2m + 1][0 = (m - 1)^2][m = 1]So, when ( m = 1 ), the line is tangent to the circle. For all other values of ( m ), the distance is less than the radius, so the line intersects the circle at two points.Therefore, depending on the value of ( m ), the line can either be tangent or intersect the circle. So, the relationship is that they either intersect or are tangent. So, option C is correct.But just to be thorough, let me plug in ( m = 1 ) into the line equation and see if it's tangent.Line equation when ( m = 1 ):[1 cdot x + y - 1 - 1 = 0][x + y - 2 = 0][y = -x + 2]Let me solve this with the circle equation ( x^2 + y^2 = 2 ).Substitute ( y = -x + 2 ) into the circle:[x^2 + (-x + 2)^2 = 2][x^2 + x^2 - 4x + 4 = 2][2x^2 - 4x + 4 - 2 = 0][2x^2 - 4x + 2 = 0][x^2 - 2x + 1 = 0][(x - 1)^2 = 0][x = 1]So, ( x = 1 ), then ( y = -1 + 2 = 1 ). So, the line touches the circle at (1,1). That's just one point, so it's tangent.For another value, say ( m = 0 ):Line equation:[0 cdot x + y - 0 - 1 = 0][y = 1]Intersection with circle ( x^2 + y^2 = 2 ):[x^2 + 1 = 2][x^2 = 1][x = pm 1]So, points (1,1) and (-1,1). Therefore, two intersection points. So, when ( m = 0 ), the line intersects the circle at two points.Therefore, depending on ( m ), the line can be tangent or intersect the circle. So, the relationship is either intersect or tangent.Hence, the correct answer is C: Intersect or Tangent.