💡To determine the area of the figure, we start by noting that each side is of equal length and each angle is a right angle. This suggests that the figure is a square or a rectangle composed of smaller squares.1. **Identify the number of smaller squares:** - The figure is divided into smaller squares, each with a side length of 2 cm. - By examining the figure, we can see that it is composed of 8 such smaller squares.2. **Calculate the area of one smaller square:** - The area of a square is given by the formula: [ text{Area} = text{side length}^2 ] - Substituting the given side length: [ text{Area of one smaller square} = 2^2 = 4 text{ cm}^2 ]3. **Calculate the total area of the figure:** - Since there are 8 smaller squares, the total area is: [ text{Total area} = 8 times 4 = 32 text{ cm}^2 ]Thus, the area of the figure is 32 square centimeters.

💡First, I need to expand the given expression: ((x^2 + 2)(3x^3 + 5x^2 + 2) - 2(x^4 - 3x^3 + 2x^2)).I'll start by expanding the product ((x^2 + 2)(3x^3 + 5x^2 + 2)):[begin{align*}(x^2 + 2)(3x^3 + 5x^2 + 2) &= x^2 cdot 3x^3 + x^2 cdot 5x^2 + x^2 cdot 2 + 2 cdot 3x^3 + 2 cdot 5x^2 + 2 cdot 2 &= 3x^5 + 5x^4 + 2x^2 + 6x^3 + 10x^2 + 4 &= 3x^5 + 5x^4 + 6x^3 + 12x^2 + 4.end{align*}]Next, I'll expand the second part of the expression, (-2(x^4 - 3x^3 + 2x^2)):[-2(x^4 - 3x^3 + 2x^2) = -2x^4 + 6x^3 - 4x^2.]Now, I'll combine the two expanded expressions:[3x^5 + 5x^4 + 6x^3 + 12x^2 + 4 - 2x^4 + 6x^3 - 4x^2.]Combining like terms:[begin{align*}3x^5 + (5x^4 - 2x^4) + (6x^3 + 6x^3) + (12x^2 - 4x^2) + 4 &= 3x^5 + 3x^4 + 12x^3 + 8x^2 + 4.end{align*}]Finally, I'll count the number of nonzero terms in the resulting polynomial:- (3x^5)- (3x^4)- (12x^3)- (8x^2)- (4)There are **5** nonzero terms.

💡First, I identify the prices of the products. Let ( S ) represent the price of a smartphone, ( PC ) represent the price of a personal computer, and ( AT ) represent the price of an advanced tablet.Given that the smartphone costs 300, I can express the personal computer's price as ( PC = S + x ), where ( x ) is the additional amount compared to the smartphone. The advanced tablet's price is the sum of the smartphone and personal computer prices, so ( AT = S + PC ).The total cost for one of each product is 2200, which leads to the equation:[S + PC + AT = 2200]Substituting the known values and expressions into the equation:[300 + (300 + x) + (300 + 300 + x) = 2200]Simplifying the equation:[300 + 300 + x + 300 + 300 + x = 2200][1200 + 2x = 2200]Solving for ( x ):[2x = 2200 - 1200][2x = 1000][x = 500]Therefore, a personal computer costs 500 more than a smartphone.

💡Okay, so I need to figure out the probability of rolling the number 1 exactly four times when I roll a fair six-sided die five times. Hmm, let me break this down step by step.First, I know that each roll of the die is independent, meaning the outcome of one roll doesn't affect the others. That's good because it simplifies things. Each face of the die has an equal chance of landing face up, so the probability of rolling a 1 on any single roll is 1 out of 6, or (frac{1}{6}). Conversely, the probability of not rolling a 1 is 5 out of 6, or (frac{5}{6}).Now, I want exactly four 1s in five rolls. That means one of the rolls will not be a 1. I need to calculate the probability for this specific scenario. Let's think about one possible sequence where the first four rolls are 1s and the fifth roll is something else. The probability for this exact sequence would be:[left(frac{1}{6}right)^4 times frac{5}{6}]But wait, the non-1 roll doesn't have to be the fifth roll specifically. It could be any one of the five rolls. So, I need to account for all the different positions where the non-1 roll could occur. How many different ways can this happen?Well, there are five rolls, and the non-1 could be in any one of those five positions. So, there are 5 different sequences where exactly four rolls are 1s and one is not. Each of these sequences has the same probability, which is (left(frac{1}{6}right)^4 times frac{5}{6}).Therefore, to get the total probability, I need to multiply the probability of one such sequence by the number of possible sequences. That would be:[5 times left(frac{1}{6}right)^4 times frac{5}{6}]Let me compute this step by step. First, calculate (left(frac{1}{6}right)^4):[left(frac{1}{6}right)^4 = frac{1}{1296}]Then, multiply this by (frac{5}{6}):[frac{1}{1296} times frac{5}{6} = frac{5}{7776}]Now, multiply this result by 5 to account for all possible positions of the non-1 roll:[5 times frac{5}{7776} = frac{25}{7776}]So, the probability of rolling exactly four 1s in five rolls is (frac{25}{7776}).Wait, let me double-check my calculations to make sure I didn't make a mistake. The number of ways to choose which roll is not a 1 is indeed 5, since there are five rolls. Each of these has the same probability, so multiplying by 5 makes sense. The probability for each specific sequence is (left(frac{1}{6}right)^4 times frac{5}{6}), which seems correct. Multiplying all together gives (frac{25}{7776}), which I believe is the right answer.I think I've covered all the necessary steps and checked my work. So, I'm confident that the probability is (frac{25}{7776}).

💡Alright, so I've got this problem here with two propositions, p and q, and I need to figure out which combination of them is true. The options are A: p ∧ q, B: ¬p ∧ q, C: p ∧ ¬q, and D: ¬p ∧ ¬q. Let me take it step by step.First, let's understand what each proposition is saying.Proposition p: For all real numbers x, x² + x + 1 is greater than 0. So, p is a universal statement claiming that no matter what real number you plug into the quadratic expression x² + x + 1, the result will always be positive.Proposition q: There exists a real number x such that x³ equals 1 minus x². So, q is an existential statement claiming that at least one real number satisfies the equation x³ = 1 - x².My goal is to determine whether p and q are true or false individually and then see which combination among the options A to D is correct.Starting with proposition p: ∀x ∈ ℝ, x² + x + 1 > 0.To check if this is true, I can analyze the quadratic expression x² + x + 1. A quadratic equation ax² + bx + c will always be positive if its discriminant is negative. The discriminant D is given by D = b² - 4ac.In this case, a = 1, b = 1, c = 1. So, D = (1)² - 4*(1)*(1) = 1 - 4 = -3.Since the discriminant is negative, the quadratic equation x² + x + 1 never touches or crosses the x-axis. That means it doesn't have any real roots, and since the coefficient of x² is positive (a = 1 > 0), the parabola opens upwards. Therefore, the quadratic expression is always positive for all real x.So, proposition p is true.Now, moving on to proposition q: ∃x ∈ ℝ, x³ = 1 - x².This is an equation: x³ + x² - 1 = 0. I need to determine if there's at least one real solution to this equation.One way to approach this is to consider the function f(x) = x³ + x² - 1 and analyze its behavior.Let's evaluate f(x) at some points to see if it changes sign, which would imply, by the Intermediate Value Theorem, that there's a root between those points.First, let's try x = 0:f(0) = 0 + 0 - 1 = -1.Next, x = 1:f(1) = 1 + 1 - 1 = 1.So, f(0) = -1 and f(1) = 1. The function changes sign between x = 0 and x = 1. Since f(x) is continuous (as it's a polynomial), by the Intermediate Value Theorem, there must be some c in (0, 1) such that f(c) = 0. Therefore, there exists a real number x where x³ = 1 - x².Hence, proposition q is also true.Now, putting it all together:p is true, q is true.So, the possible options are:A: p ∧ q – both p and q are true.B: ¬p ∧ q – p is false and q is true.C: p ∧ ¬q – p is true and q is false.D: ¬p ∧ ¬q – both p and q are false.Since both p and q are true, the correct option is A: p ∧ q.I don't think I made any mistakes here. For p, the discriminant method is solid for quadratics, and for q, the Intermediate Value Theorem application seems correct. So, I'm confident that A is the right answer.**Final Answer**boxed{A}

💡First, I need to determine the number of students taking only geometry. Since there are 40 students taking geometry in total and 15 of them are also taking biology, I subtract the 15 from 40 to find the number of students taking only geometry.Next, I know there are 20 students taking only biology.Finally, to find the total number of students taking either geometry or biology but not both, I add the number of students taking only geometry to the number of students taking only biology.