💡First, I need to determine how many minutes Jenna can tan in the first two weeks of the month. She has a total monthly limit of 200 minutes and plans to use 80 minutes in the last two weeks. Subtracting the last two weeks' tanning time from the total monthly allowance gives:200 minutes - 80 minutes = 120 minutes for the first two weeks.Next, I'll calculate how many days a week Jenna tans during the first two weeks. Let’s assume she tans ( x ) days a week. Over two weeks, she would tan ( 2x ) days.Since she tans 30 minutes each day, the total tanning time for ( 2x ) days is:30 minutes/day * 2x days = 60x minutes.Setting this equal to the 120 minutes she has for the first two weeks:60x = 120.Solving for ( x ):x = 120 / 60x = 2.Therefore, Jenna tans 2 days a week during the first two weeks.

💡Alright, so I've got this math problem here, and I need to figure out the value of this expression:[sqrt{1} + sqrt{1+2} + sqrt{1+2+3} + sqrt{1+2+3+4}]And then choose the correct option from A to E. Let me break this down step by step because I want to make sure I understand each part correctly.First, let's look at each term inside the square roots individually. The expression has four terms:1. (sqrt{1})2. (sqrt{1+2})3. (sqrt{1+2+3})4. (sqrt{1+2+3+4})Okay, so I need to simplify each of these square roots one by one.Starting with the first term, (sqrt{1}). That's straightforward because the square root of 1 is just 1. So, that part is easy.Next, the second term is (sqrt{1+2}). Let me compute what's inside the square root first: 1 plus 2 equals 3. So, this simplifies to (sqrt{3}). I remember that (sqrt{3}) is an irrational number, approximately 1.732, but since the question asks for the simplest form, I'll just keep it as (sqrt{3}).Moving on to the third term: (sqrt{1+2+3}). Again, let's add the numbers inside the square root first. 1 plus 2 is 3, and then 3 plus 3 is 6. So, this term simplifies to (sqrt{6}). Similar to the previous one, (sqrt{6}) is also irrational, approximately 2.449, but I'll keep it as (sqrt{6}) for the simplest form.Now, the fourth term is (sqrt{1+2+3+4}). Let's add those numbers: 1 plus 2 is 3, plus 3 is 6, and plus 4 is 10. So, this term simplifies to (sqrt{10}). Again, (sqrt{10}) is irrational, roughly 3.162, but I'll stick with (sqrt{10}) for simplicity.So, after simplifying each term, the entire expression becomes:[sqrt{1} + sqrt{3} + sqrt{6} + sqrt{10}]Which is:[1 + sqrt{3} + sqrt{6} + sqrt{10}]Now, let's look at the options provided:A) (1 + 2sqrt{3} + sqrt{6} + sqrt{10})B) (1 + sqrt{3} + sqrt{6} + sqrt{10})C) (2 + sqrt{3} + sqrt{6} + 2sqrt{10})D) (4 + sqrt{3} + sqrt{6} + sqrt{10})E) (1 + 3sqrt{3} + 2sqrt{6} + sqrt{10})Comparing my simplified expression to these options, it matches option B exactly. So, option B is the correct answer.But wait, just to make sure I didn't make any mistakes, let me double-check each step.First, (sqrt{1}) is definitely 1. Then, (sqrt{1+2}) is (sqrt{3}), that's correct. Next, (sqrt{1+2+3}) is (sqrt{6}), which is accurate. Finally, (sqrt{1+2+3+4}) is (sqrt{10}), that's right too.Adding them all together: 1 + (sqrt{3}) + (sqrt{6}) + (sqrt{10}), which is exactly option B. So, I'm confident that option B is the correct answer.I was a bit confused at first because I thought maybe there was a trick or a pattern I was missing, but it seems straightforward once I break it down term by term. I guess sometimes these problems look complicated, but they're just about carefully simplifying each part.Another thing I considered was whether any of the square roots could be simplified further, but (sqrt{3}), (sqrt{6}), and (sqrt{10}) are all in their simplest forms because they don't have any perfect square factors other than 1. So, there's no need to simplify them further.I also thought about whether the problem might be asking for a numerical approximation, but since the options are all in terms of square roots, it's clear that the answer should be in its exact form, not a decimal approximation.In summary, by simplifying each term individually and then adding them together, I arrived at the conclusion that the correct answer is option B: (1 + sqrt{3} + sqrt{6} + sqrt{10}).

💡First, I need to determine the total number of students who prefer chocolate, apple, or blueberry pie by adding the given numbers: 15 (chocolate) + 10 (apple) + 9 (blueberry) = 34 students.Next, I'll find out how many students are left who prefer other types of pies by subtracting this total from the overall number of students: 48 - 34 = 14 students.Out of these 14 students, one-third prefer cherry pie. Calculating one-third of 14 gives approximately 4.67 students. Since the number of students must be a whole number, I'll round it to 5 students.Finally, to determine the degrees for cherry pie in the pie chart, I'll use the formula: (5 students / 48 total students) × 360 degrees = 37.5 degrees.

💡First, I need to determine the total number of pizza slices Mary initially had. Since each large pizza has 8 slices and she ordered 2 large pizzas, the total number of slices is 2 multiplied by 8, which equals 16 slices.Mary has 9 slices remaining after eating some. To find out how many slices she ate, I subtract the remaining slices from the initial total. So, 16 slices minus 9 remaining slices equals 7 slices.Therefore, Mary ate 7 slices.

💡First, I need to find the sum of vectors (overrightarrow{a}) and (overrightarrow{b}), as well as twice vector (overrightarrow{a}) minus vector (overrightarrow{b}).Calculating (overrightarrow{a} + overrightarrow{b}):[overrightarrow{a} + overrightarrow{b} = (2, 1) + (1, x) = (3, 1 + x)]Calculating (2overrightarrow{a} - overrightarrow{b}):[2overrightarrow{a} - overrightarrow{b} = 2(2, 1) - (1, x) = (4, 2) - (1, x) = (3, 2 - x)]Since (overrightarrow{a} + overrightarrow{b}) is parallel to (2overrightarrow{a} - overrightarrow{b}), their corresponding components must be proportional. This gives the equation:[frac{3}{3} = frac{1 + x}{2 - x}]Simplifying the equation:[1 = frac{1 + x}{2 - x}]Cross-multiplying to solve for (x):[2 - x = 1 + x][2x = 1][x = frac{1}{2}]

💡Okay, so I have two trigonometry problems to solve here. Let me take them one at a time. Starting with the first one: Simplify the expression (dfrac{cos (α - dfrac{π}{2})}{sin (dfrac{5π}{2} + α)} cdot sin (α - 2π) cdot cos (2π - α)). Hmm, that looks a bit complicated, but maybe I can break it down using some trigonometric identities.First, I remember that cosine and sine have some co-function identities. Specifically, (cos(θ - dfrac{π}{2})) is equal to (sin θ). Let me verify that. Yes, because shifting cosine by π/2 gives sine. So, (cos(α - dfrac{π}{2}) = sin α). That simplifies the numerator.Now, looking at the denominator: (sin(dfrac{5π}{2} + α)). I know that sine has a period of (2π), so adding (2π) to the angle doesn't change its value. Let me subtract (2π) from (dfrac{5π}{2}) to make it simpler. (dfrac{5π}{2} - 2π = dfrac{5π}{2} - dfrac{4π}{2} = dfrac{π}{2}). So, (sin(dfrac{5π}{2} + α) = sin(dfrac{π}{2} + α)). I also remember that (sin(dfrac{π}{2} + α) = cos α). Let me double-check that identity. Yes, because sine of π/2 plus alpha is cosine alpha. So, the denominator simplifies to (cos α).So now, the first fraction simplifies to (dfrac{sin α}{cos α}), which is just (tan α). Okay, that's simpler.Next, looking at the second part of the expression: (sin(α - 2π)). Sine has a period of (2π), so subtracting (2π) from the angle doesn't change its value. Therefore, (sin(α - 2π) = sin α).Then, the third part: (cos(2π - α)). I recall that cosine is an even function, so (cos(2π - α) = cos α). Let me confirm that. Yes, because cosine is symmetric around 0, so subtracting alpha from 2π gives the same value as cosine alpha.Putting it all together, the entire expression becomes:[tan α cdot sin α cdot cos α]But (tan α) is (dfrac{sin α}{cos α}), so substituting that in:[dfrac{sin α}{cos α} cdot sin α cdot cos α]I can see that the (cos α) in the denominator and the (cos α) in the numerator will cancel each other out. So, that leaves me with:[sin α cdot sin α = sin^2 α]So, the simplified form of the expression is (sin^2 α). That wasn't too bad once I broke it down step by step.Moving on to the second problem: Evaluate (sin dfrac{25π}{6} + cos dfrac{25π}{3} + tan (-dfrac{25π}{4})). Okay, so I need to evaluate each of these trigonometric functions separately and then add them up.Starting with (sin dfrac{25π}{6}). I know that sine has a period of (2π), so I can subtract multiples of (2π) to find an equivalent angle between 0 and (2π). Let me calculate how many times (2π) goes into (dfrac{25π}{6}).First, (2π = dfrac{12π}{6}). So, dividing (dfrac{25π}{6}) by (dfrac{12π}{6}) gives me approximately 2.0833. So, that means I can subtract (2π) twice from (dfrac{25π}{6}).Calculating (25π/6 - 2π = 25π/6 - 12π/6 = 13π/6). But 13π/6 is still more than 2π, so let me subtract another 2π: 13π/6 - 12π/6 = π/6. So, (sin dfrac{25π}{6} = sin dfrac{π}{6}).I remember that (sin dfrac{π}{6} = dfrac{1}{2}). So, that part is straightforward.Next, (cos dfrac{25π}{3}). Cosine also has a period of (2π), so I'll do the same process. Let me find how many times (2π) goes into (dfrac{25π}{3}).(2π = dfrac{6π}{3}). Dividing (dfrac{25π}{3}) by (dfrac{6π}{3}) gives approximately 4.1667. So, I can subtract (2π) four times from (dfrac{25π}{3}).Calculating (25π/3 - 4*(2π) = 25π/3 - 8π = 25π/3 - 24π/3 = π/3). So, (cos dfrac{25π}{3} = cos dfrac{π}{3}).I remember that (cos dfrac{π}{3} = dfrac{1}{2}). So, that part is also straightforward.Now, the last term: (tan (-dfrac{25π}{4})). Tangent has a period of (π), so I can add multiples of (π) to find an equivalent angle between 0 and (π). Also, tangent is an odd function, so (tan(-θ) = -tan θ). So, (tan (-dfrac{25π}{4}) = -tan dfrac{25π}{4}).Let me simplify (dfrac{25π}{4}). Since the period is (π), I can subtract multiples of (π) to find an equivalent angle. Let's see, (25π/4 - 6π = 25π/4 - 24π/4 = π/4). So, (tan dfrac{25π}{4} = tan dfrac{π}{4}).I know that (tan dfrac{π}{4} = 1). Therefore, (tan (-dfrac{25π}{4}) = -1).Putting it all together, the expression becomes:[sin dfrac{25π}{6} + cos dfrac{25π}{3} + tan (-dfrac{25π}{4}) = dfrac{1}{2} + dfrac{1}{2} - 1]Adding those up: (dfrac{1}{2} + dfrac{1}{2} = 1), and then (1 - 1 = 0). So, the value of the expression is 0.Wait, let me double-check my steps to make sure I didn't make any mistakes. For the first problem, I used co-function identities and periodicity correctly, and everything canceled out nicely to give (sin^2 α). For the second problem, I reduced each angle by subtracting multiples of their periods, applied the appropriate identities, and the arithmetic checks out. So, I think both answers are correct.