💡First, I need to determine the age of the rabbit. Since the rabbit is half the age of the cat, and the cat is 8 years old, the rabbit's age is 8 divided by 2, which equals 4 years.Next, I calculate the age of the dog. The dog is three times as old as the rabbit. Given that the rabbit is 4 years old, the dog's age is 4 multiplied by 3, resulting in 12 years.Therefore, the dog is 12 years old.

💡First, identify the vowels and consonants. Vowels are A, E, I, O, U, and Y, making a total of 6 vowels. Since W is considered a consonant, there are 26 total letters minus 6 vowels, resulting in 20 consonants. The probability of selecting a student with consonant initials is the number of consonants divided by the total number of letters, which is 20/26. Simplifying this fraction gives 10/13.

💡First, I need to determine the value of (sin 300^{circ}). I recall that (300^{circ}) can be expressed as (360^{circ} - 60^{circ}). This allows me to use the cofunction identity for sine.Using the identity (sin(360^{circ} - theta) = -sin theta), I substitute (theta = 60^{circ}):[sin 300^{circ} = sin(360^{circ} - 60^{circ}) = -sin 60^{circ}]Next, I know that (sin 60^{circ} = dfrac{sqrt{3}}{2}). Therefore:[sin 300^{circ} = -dfrac{sqrt{3}}{2}]So, the correct answer is (-dfrac{sqrt{3}}{2}).

💡Okay, so I have this quadratic function: y = ax² - 3x + a² - 1. The problem says that its graph opens upwards and passes through the origin, which is (0,0). I need to find the value of 'a'. Alright, let's break this down step by step.First, I remember that for a quadratic function in the form y = ax² + bx + c, the coefficient 'a' determines the direction in which the parabola opens. If 'a' is positive, it opens upwards, and if 'a' is negative, it opens downwards. Since the problem states that the graph opens upwards, that means 'a' must be greater than zero. So, I can note that a > 0.Next, the graph passes through the origin, which is the point (0,0). That means when x is 0, y is also 0. So, I can substitute x = 0 and y = 0 into the equation to find a relationship involving 'a'.Let's do that substitution:0 = a*(0)² - 3*(0) + a² - 1Simplifying this, the terms with x in them become zero:0 = 0 - 0 + a² - 1So, this simplifies to:0 = a² - 1Now, I need to solve for 'a'. Let's rearrange this equation:a² - 1 = 0This is a simple quadratic equation. I can solve for 'a' by isolating it. Let's add 1 to both sides:a² = 1Now, to find 'a', I take the square root of both sides:a = ±√1Which gives:a = 1 or a = -1But wait, earlier I noted that since the parabola opens upwards, 'a' must be positive. So, a = -1 is not a valid solution in this context because it would make the parabola open downwards.Therefore, the only valid solution is:a = 1Let me double-check to make sure I didn't make any mistakes. If a = 1, then the function becomes:y = 1*x² - 3x + (1)² - 1y = x² - 3x + 1 - 1y = x² - 3xDoes this pass through the origin? Let's plug in x = 0:y = 0² - 3*0 = 0Yes, it does pass through (0,0). Also, since the coefficient of x² is positive (1), the parabola opens upwards, which matches the given condition.So, everything checks out. The value of 'a' is indeed 1.

💡Okay, so I need to figure out how many matchsticks Farrah ordered. Let's see, she ordered 4 boxes from Amazon, and each box has 20 matchboxes. Each matchbox has 300 sticks. Hmm, so I guess I need to multiply these numbers together to get the total number of sticks.First, let's think about the boxes. She has 4 boxes, and each box has 20 matchboxes. So, to find out how many matchboxes she has in total, I should multiply 4 by 20. That should give me the total number of matchboxes.4 times 20 is 80. So, Farrah has 80 matchboxes in total.Now, each matchbox has 300 sticks. So, to find out how many sticks she has altogether, I need to multiply the number of matchboxes by the number of sticks in each matchbox. That means I need to multiply 80 by 300.Let me do that multiplication. 80 times 300. Well, 80 times 3 is 240, and since there are two zeros in 300, I need to add two zeros to 240, which gives me 24,000.So, putting it all together, Farrah ordered 4 boxes, each with 20 matchboxes, and each matchbox has 300 sticks. Multiplying these together, 4 times 20 is 80 matchboxes, and 80 times 300 is 24,000 sticks.I think that's the total number of matchsticks Farrah ordered.

💡Okay, so I have this problem where I need to find the first and second derivatives of an implicit function given by the equation:ln sqrt{x^{2}+y^{2}} - operatorname{arctg} frac{y}{x} = 0Hmm, implicit functions can be tricky, but I remember that to find derivatives, I can use implicit differentiation. Let me try to recall the steps involved.First, I think I need to differentiate both sides of the equation with respect to ( x ). Since ( y ) is a function of ( x ), I'll have to apply the chain rule wherever there's a ( y ).Let me rewrite the equation for clarity:ln sqrt{x^{2} + y^{2}} - arctan left( frac{y}{x} right) = 0Alright, let's differentiate term by term.Starting with the first term: ( ln sqrt{x^{2} + y^{2}} ). I know that ( ln sqrt{A} ) is the same as ( frac{1}{2} ln A ), so maybe that simplifies things a bit. Let me write it as:frac{1}{2} ln (x^{2} + y^{2})Now, differentiating this with respect to ( x ):The derivative of ( ln (x^{2} + y^{2}) ) with respect to ( x ) is ( frac{1}{x^{2} + y^{2}} ) times the derivative of the inside, which is ( 2x + 2y frac{dy}{dx} ). But since we have a ( frac{1}{2} ) factor, it becomes:frac{1}{2} times frac{2x + 2y frac{dy}{dx}}{x^{2} + y^{2}} = frac{x + y frac{dy}{dx}}{x^{2} + y^{2}}Okay, that's the derivative of the first term.Now, the second term is ( -arctan left( frac{y}{x} right) ). The derivative of ( arctan(u) ) with respect to ( x ) is ( frac{1}{1 + u^2} times frac{du}{dx} ). So here, ( u = frac{y}{x} ), so ( frac{du}{dx} = frac{x frac{dy}{dx} - y}{x^2} ).Putting it all together, the derivative of the second term is:- left( frac{1}{1 + left( frac{y}{x} right)^2} times frac{x frac{dy}{dx} - y}{x^2} right)Simplify ( 1 + left( frac{y}{x} right)^2 ) to ( frac{x^2 + y^2}{x^2} ), so the expression becomes:- left( frac{x^2}{x^2 + y^2} times frac{x frac{dy}{dx} - y}{x^2} right) = - frac{x frac{dy}{dx} - y}{x^2 + y^2}So, combining both derivatives, we have:frac{x + y frac{dy}{dx}}{x^{2} + y^{2}} - frac{x frac{dy}{dx} - y}{x^2 + y^2} = 0Since both terms have the same denominator, I can combine them:frac{x + y frac{dy}{dx} - x frac{dy}{dx} + y}{x^{2} + y^{2}} = 0Simplify the numerator:x + y frac{dy}{dx} - x frac{dy}{dx} + y = (x + y) + frac{dy}{dx}(y - x)So, the equation becomes:frac{(x + y) + frac{dy}{dx}(y - x)}{x^{2} + y^{2}} = 0Since the denominator ( x^{2} + y^{2} ) is never zero (assuming ( x ) and ( y ) are not both zero), we can set the numerator equal to zero:(x + y) + frac{dy}{dx}(y - x) = 0Now, solve for ( frac{dy}{dx} ):frac{dy}{dx}(y - x) = - (x + y)So,frac{dy}{dx} = - frac{x + y}{y - x} = frac{x + y}{x - y}Alright, that's the first derivative. Now, I need to find the second derivative ( frac{d^2y}{dx^2} ).To find ( y'' ), I'll differentiate ( y' = frac{x + y}{x - y} ) with respect to ( x ). Let's use the quotient rule for differentiation.Let me denote ( u = x + y ) and ( v = x - y ), so ( y' = frac{u}{v} ).The quotient rule states that:left( frac{u}{v} right)' = frac{u'v - uv'}{v^2}So, I need to find ( u' ) and ( v' ).First, ( u = x + y ), so:u' = 1 + y'Similarly, ( v = x - y ), so:v' = 1 - y'Now, plug these into the quotient rule:y'' = frac{(1 + y')(x - y) - (x + y)(1 - y')}{(x - y)^2}Let me expand the numerator:First term: ( (1 + y')(x - y) = (x - y) + y'(x - y) )Second term: ( (x + y)(1 - y') = (x + y) - y'(x + y) )So, the numerator becomes:(x - y) + y'(x - y) - (x + y) + y'(x + y)Simplify term by term:1. ( (x - y) - (x + y) = x - y - x - y = -2y )2. ( y'(x - y) + y'(x + y) = y'(x - y + x + y) = y'(2x) )So, the numerator simplifies to:-2y + 2x y'Therefore, the second derivative is:y'' = frac{-2y + 2x y'}{(x - y)^2}But we already have ( y' = frac{x + y}{x - y} ), so let's substitute that in:y'' = frac{-2y + 2x left( frac{x + y}{x - y} right)}{(x - y)^2}Let me factor out the 2 in the numerator:y'' = frac{2 left( -y + x left( frac{x + y}{x - y} right) right)}{(x - y)^2}Simplify the expression inside the parentheses:-y + x left( frac{x + y}{x - y} right) = -y left( frac{x - y}{x - y} right) + frac{x(x + y)}{x - y}So,= frac{ -y(x - y) + x(x + y) }{x - y}Expand the numerator:- yx + y^2 + x^2 + xy = x^2 + y^2So, the expression inside the parentheses simplifies to:frac{x^2 + y^2}{x - y}Therefore, the second derivative becomes:y'' = frac{2 left( frac{x^2 + y^2}{x - y} right)}{(x - y)^2} = frac{2(x^2 + y^2)}{(x - y)^3}So, putting it all together, the first derivative is ( frac{dy}{dx} = frac{x + y}{x - y} ) and the second derivative is ( frac{d^2y}{dx^2} = frac{2(x^2 + y^2)}{(x - y)^3} ).I think that's it. Let me just double-check my steps to make sure I didn't make any mistakes.Starting from the original equation, I used implicit differentiation correctly, applied the chain rule and the quotient rule appropriately. The algebra steps seem to check out, especially when simplifying the numerator for the second derivative. I also correctly substituted ( y' ) into the expression for ( y'' ).So, I feel confident that these are the correct first and second derivatives.