Example Question - formula
Here are examples of questions we've helped users solve.
Finding the Inverse of an Exponential Function
<p>Let \( y = e^{4x} - 5 \)</p> <p>Swap x and y to find the inverse: \( x = e^{4y} - 5 \)</p> <p>Add 5 to both sides: \( x + 5 = e^{4y} \)</p> <p>Take the natural logarithm of both sides: \( \ln(x + 5) = \ln(e^{4y}) \)</p> <p>Use the property of logarithms: \( \ln(x + 5) = 4y \)</p> <p>Divide by 4: \( y = \frac{1}{4}\ln(x + 5) \)</p> <p>The inverse function is \( f^{-1}(x) = \frac{1}{4}\ln(x + 5) \)</p>
Calculating Intersection Point of Diagonals in Rectangle
在這個問題中，我們需要確定矩形PQRS中對角線PT和QS的交點T的長度TS。 首先，我們可以觀察到，因為PQRS是矩形，所以對角線PS和QT將等長且交於中點。由此，TS將是PS的一半。 要找出PS的長度，我們可以使用畢氏定理，因為PQ和QR是矩形的兩個直角邊: \(PS^2 = PQ^2 + QR^2\) 代入已知的長度: \(PS^2 = 24^2 + 12^2 = 576 + 144 = 720\) 然後，開平方根計算PS: \(PS = \sqrt{720}\) 這給了我們PS的長度。為了獲得TS，我們只需將結果除以2： \(TS = \frac{PS}{2} = \frac{\sqrt{720}}{2}\) 現在來計算它的確切值: \(PS = \sqrt{720} \approx 26.83 \, \text{cm}\) (保留兩位小數) 所以, \(TS = \frac{26.83}{2} \approx 13.42 \, \text{cm}\) (保留兩位小數) 所以TS的長度大約是13.42厘米。
Solving Quadratic Equations with Formula
Claro, resolveremos la ecuación cuadrática que aparece en la imagen, que es \(2x^2 + 5x + 3 = 0\). Para resolver esta ecuación, podemos utilizar la fórmula cuadrática: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Donde \(a\), \(b\), y \(c\) son los coeficientes de los términos cuadráticos, lineales e independientes, respectivamente, de nuestra ecuación. En nuestra ecuación, \(a = 2\), \(b = 5\), y \(c = 3\). Primero calculamos el discriminante (\(b^2 - 4ac\)): \[\begin{align*} b^2 - 4ac &= 5^2 - 4(2)(3) \\ &= 25 - 24 \\ &= 1 \end{align*}\] Como el discriminante es positivo, habrá dos soluciones reales y distintas. Ahora utilizamos la fórmula cuadrática para encontrar los valores de \(x\): \[ x = \frac{-5 \pm \sqrt{1}}{2(2)} = \frac{-5 \pm 1}{4} \] Ahora calculamos los dos valores posibles para \(x\): 1) Tomando el signo positivo: \[ x = \frac{-5 + 1}{4} = \frac{-4}{4} = -1 \] 2) Tomando el signo negativo: \[ x = \frac{-5 - 1}{4} = \frac{-6}{4} = -\frac{3}{2} = -1.5 \] Entonces, las dos soluciones para la ecuación cuadrática son \(x = -1\) y \(x = -1.5\).
Solving for n in a Binomial Test
The image presents a question related to a binomial test: For a binomial test, if p = 1/2, p*n = 35 and q*n = 35, what is the value of n? Here, p is the probability of success on a single trial, and q is the probability of failure (q = 1 - p). Since we know p, we can find q: q = 1 - p = 1 - 1/2 = 1/2 With the formulas p*n and q*n equaling 35, we have: p*n = 35 q*n = 35 Since both p and q are equal to 1/2, and both p*n and q*n are equal to 35, we can solve for n: (1/2)*n = 35 n = 35 / (1/2) n = 35 * 2 n = 70 Thus, the value of n is 70.
Matrix Multiplication Result
The image shows two matrices that need to be multiplied. Let's call the first matrix A and the second matrix B: Matrix A = [3 2 5] [2 3 1] Matrix B = [4 5 -5] [5 -1 6] [4 5 -5] To multiply two matrices, we can use the following formula for the elements of the resulting matrix: Given matrices A (of size m by n) and B (of size n by p), their product C (of size m by p) will have elements: \[ c_{ij} = \sum_{k=1}^{n} a_{ik} \cdot b_{kj} \] So for matrices A (2x3) and B (3x3), we can find the product C (2x3) by multiplying each row of A with each column of B. Let's perform the multiplication: \[ c_{11} = (3 \times 4) + (2 \times 5) + (5 \times 4) = 12 + 10 + 20 = 42 \] \[ c_{12} = (3 \times 5) + (2 \times -1) + (5 \times 5) = 15 - 2 + 25 = 38 \] \[ c_{13} = (3 \times -5) + (2 \times 6) + (5 \times -5) = -15 + 12 - 25 = -28 \] \[ c_{21} = (2 \times 4) + (3 \times 5) + (1 \times 4) = 8 + 15 + 4 = 27 \] \[ c_{22} = (2 \times 5) + (3 \times -1) + (1 \times 5) = 10 - 3 + 5 = 12 \] \[ c_{23} = (2 \times -5) + (3 \times 6) + (1 \times -5) = -10 + 18 - 5 = 3 \] So the product C of the matrices A and B is: Matrix C = [42 38 -28] [27 12 3]
Calculating Continuous Compounding Interest
To solve this problem, we use the formula for continuous compounding interest: A = P * e^(rt) Where: A = the amount of money accumulated after n years, including interest. P = the principal amount (the initial amount of money) r = the annual interest rate (decimal) t = the time the money is invested for, in years e = the mathematical constant approximately equal to 2.71828 Given: P = $1,800 r = 3.7% per annum = 0.037 (as a decimal) t = 10 years We can now substitute the given values into the formula to find the value of the account after 10 years: A = 1800 * e^(0.037 * 10) Calculating the exponent first: 0.037 * 10 = 0.37 Now, raise e to the power of 0.37: e^0.37 is approximately 1.4481 Now multiply this by the principal amount: A = 1800 * 1.4481 ≈ 2606.58 Rounding to the nearest dollar, we get A ≈ $2,607. Therefore, the correct answer, rounded to the nearest dollar, is about $2,607 which corresponds to option C.
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