💡To determine the equation of the resulting parabola after shifting the original parabola ( y = x^2 + 1 ) first 2 units to the left and then 3 units downward, follow these steps:1. **Shift 2 units to the left**: - Shifting a function ( f(x) ) to the left by ( h ) units involves replacing ( x ) with ( x + h ). - For this problem, ( h = 2 ), so the equation becomes: [ y = (x + 2)^2 + 1 ]2. **Shift 3 units downward**: - Shifting a function ( f(x) ) downward by ( k ) units involves subtracting ( k ) from the entire function. - For this problem, ( k = 3 ), so the equation becomes: [ y = (x + 2)^2 + 1 - 3 ] Simplifying this: [ y = (x + 2)^2 - 2 ]Therefore, the equation of the resulting parabola is ( y = (x + 2)^2 - 2 ).

💡First, I need to determine how much was given to the first set of homeless families.The total amount spent is 900.00.She gave 260.00 to the second set of families and 315.00 to the last set of families.To find the amount given to the first set, I will subtract the amounts given to the second and third sets from the total amount.So, 900.00 minus 260.00 equals 640.00.Then, 640.00 minus 315.00 equals 325.00.Therefore, she gave 325.00 to the first set of homeless families.

💡Okay, so I have this equation: sin x + cos x = 1, and I need to find all the values of x between 0 and 2π that satisfy it. Hmm, let me think about how to approach this.First, I remember that sometimes when you have both sine and cosine in an equation, it can be helpful to use some trigonometric identities to simplify it. Maybe I can square both sides to use the Pythagorean identity? Let me try that.If I square both sides, I get:(sin x + cos x)^2 = 1^2Which expands to:sin²x + 2 sinx cosx + cos²x = 1Now, I know that sin²x + cos²x = 1, so I can substitute that in:1 + 2 sinx cosx = 1Subtracting 1 from both sides gives:2 sinx cosx = 0Hmm, so 2 sinx cosx equals zero. I remember that 2 sinx cosx is the same as sin(2x), so this simplifies to:sin(2x) = 0Alright, so when does sin(2x) equal zero? Well, sine of an angle is zero at multiples of π. So, 2x = kπ, where k is any integer.Solving for x, we get:x = (kπ)/2Now, since x has to be between 0 and 2π, let's find all the possible values of k that satisfy this condition.If k = 0, then x = 0If k = 1, then x = π/2If k = 2, then x = πIf k = 3, then x = 3π/2If k = 4, then x = 2π, but since the interval is 0 ≤ x < 2π, 2π is not included.So, the possible solutions are x = 0, π/2, π, and 3π/2.But wait, I squared both sides of the equation earlier, which can sometimes introduce extraneous solutions. I need to check each of these solutions in the original equation to make sure they actually work.Let's test x = 0:sin(0) + cos(0) = 0 + 1 = 1. That works.x = π/2:sin(π/2) + cos(π/2) = 1 + 0 = 1. That also works.x = π:sin(π) + cos(π) = 0 + (-1) = -1. That doesn't equal 1, so x = π is not a solution.x = 3π/2:sin(3π/2) + cos(3π/2) = -1 + 0 = -1. Again, that doesn't equal 1, so x = 3π/2 is also not a solution.So, after checking, the only solutions are x = 0 and x = π/2.Wait, but x = 0 is included in the interval, right? The problem says 0 ≤ x < 2π, so x = 0 is allowed. Okay, so that's fine.Is there another way to approach this problem without squaring both sides? Maybe using a different identity or method? Let me think.I remember that sinx + cosx can be rewritten using the sine addition formula. Specifically, sinx + cosx can be expressed as √2 sin(x + π/4). Let me verify that.Using the identity: a sinx + b cosx = √(a² + b²) sin(x + φ), where φ = arctan(b/a). In this case, a = 1 and b = 1, so φ = arctan(1/1) = π/4. And √(1² + 1²) = √2. So, yes, sinx + cosx = √2 sin(x + π/4).So, the equation becomes:√2 sin(x + π/4) = 1Dividing both sides by √2:sin(x + π/4) = 1/√2Now, when does sinθ = 1/√2? That occurs at θ = π/4 + 2πk and θ = 3π/4 + 2πk, where k is any integer.So, setting θ = x + π/4, we have:x + π/4 = π/4 + 2πk or x + π/4 = 3π/4 + 2πkSolving for x in both cases:First case:x + π/4 = π/4 + 2πkSubtract π/4 from both sides:x = 0 + 2πkSecond case:x + π/4 = 3π/4 + 2πkSubtract π/4 from both sides:x = π/2 + 2πkNow, considering 0 ≤ x < 2π, let's find the specific solutions.For the first case, x = 0 + 2πk:When k = 0, x = 0When k = 1, x = 2π, which is not included since x < 2π.For the second case, x = π/2 + 2πk:When k = 0, x = π/2When k = 1, x = π/2 + 2π, which is beyond 2π.So, the solutions are x = 0 and x = π/2, which matches what I found earlier.Therefore, both methods lead me to the same conclusion. It's good to verify using different approaches to make sure the solutions are correct.I think I've covered all the necessary steps and checked my work. It seems like x = 0 and x = π/2 are the only solutions in the given interval.**Final Answer**The solutions are boxed{0} and boxed{dfrac{pi}{2}}.

💡First, I need to determine the taxable income by subtracting the non-taxable portion from the total income. The total income is 20,000 yuan, and the non-taxable portion is 5,000 yuan. 20,000 yuan (total income) - 5,000 yuan (non-taxable) = 15,000 yuan (taxable income)Next, I need to apply the tax rates to the appropriate portions of the taxable income based on the provided tax brackets.1. For the first 3,000 yuan of taxable income, the tax rate is 3%. 3,000 yuan * 3% = 90 yuan2. For the next 9,000 yuan (from 3,001 to 12,000 yuan), the tax rate is 10%. 9,000 yuan * 10% = 900 yuan3. For the remaining 3,000 yuan (from 12,001 to 15,000 yuan), the tax rate is 20%. 3,000 yuan * 20% = 600 yuanFinally, I sum up the taxes from each portion to find the total tax amount.90 yuan + 900 yuan + 600 yuan = 1,590 yuanTherefore, the total amount of personal income tax the employee should pay in August is 1,590 yuan.

💡Okay, so I have this problem where I need to find the value of ( x^2 + y^2 ) given that ( x - y = 25 ) and ( xy = 36 ). Additionally, I need to find what ( x + y ) equals. Hmm, let me think about how to approach this.First, I remember that there are some algebraic identities that relate ( x + y ), ( x - y ), and ( xy ) to ( x^2 + y^2 ). Specifically, I recall that:[(x - y)^2 = x^2 - 2xy + y^2]And also:[(x + y)^2 = x^2 + 2xy + y^2]From these, I can see that if I can find ( (x + y)^2 ) or ( (x - y)^2 ), I might be able to find ( x^2 + y^2 ). Since I already know ( x - y = 25 ), maybe I can use that.Let me write down what I know:1. ( x - y = 25 )2. ( xy = 36 )I need to find ( x^2 + y^2 ) and ( x + y ).Starting with ( x - y = 25 ), if I square both sides, I get:[(x - y)^2 = 25^2][x^2 - 2xy + y^2 = 625]But I also know that ( xy = 36 ), so I can substitute that into the equation:[x^2 - 2(36) + y^2 = 625][x^2 - 72 + y^2 = 625]Now, I can solve for ( x^2 + y^2 ):[x^2 + y^2 = 625 + 72][x^2 + y^2 = 697]Okay, so that gives me ( x^2 + y^2 = 697 ). Now, I need to find ( x + y ). I remember that:[(x + y)^2 = x^2 + 2xy + y^2]I already know ( x^2 + y^2 = 697 ) and ( xy = 36 ), so I can substitute those values in:[(x + y)^2 = 697 + 2(36)][(x + y)^2 = 697 + 72][(x + y)^2 = 769]To find ( x + y ), I take the square root of both sides:[x + y = sqrt{769}]Wait, but square roots can be positive or negative. Should I consider both possibilities? Hmm, in the original problem, there's no information about whether ( x ) and ( y ) are positive or negative. However, since ( x - y = 25 ) is positive, and ( xy = 36 ) is positive, it suggests that both ( x ) and ( y ) are positive numbers because their product is positive and their difference is positive. So, ( x ) must be larger than ( y ), and both are positive. Therefore, ( x + y ) should also be positive.So, ( x + y = sqrt{769} ). Let me double-check my calculations to make sure I didn't make any mistakes.Starting with ( x - y = 25 ), squaring gives ( 625 ). Then, ( x^2 + y^2 = 625 + 2xy ). Since ( xy = 36 ), that's ( 625 + 72 = 697 ). That seems correct.Then, for ( x + y ), using ( (x + y)^2 = x^2 + y^2 + 2xy = 697 + 72 = 769 ). Taking the square root gives ( sqrt{769} ). That also seems correct.I think I've got it right. So, ( x^2 + y^2 = 697 ) and ( x + y = sqrt{769} ).

💡First, I'll denote the present ages of p and q as ( p ) and ( q ) respectively.From the problem, 12 years ago, p was half the age of q. This gives me the equation:[p - 12 = frac{1}{2}(q - 12)]I also know that the total of their present ages is 42:[p + q = 42]Next, I'll solve the first equation for ( p ):[p = frac{1}{2}q - 6 + 12][p = frac{1}{2}q + 6]Substituting this expression for ( p ) into the second equation:[left(frac{1}{2}q + 6right) + q = 42][frac{3}{2}q + 6 = 42][frac{3}{2}q = 36][q = 24]Now, substituting ( q = 24 ) back into the equation ( p + q = 42 ):[p + 24 = 42][p = 18]Finally, the ratio of their present ages ( p : q ) is:[18 : 24 = 3 : 4]