Example Question - solving for y
Here are examples of questions we've helped users solve.
Finding Inverse of Exponential Function
To find the inverse of the exponential function \( y = 4^x \), you need to solve for x in terms of y. Starting with the original function: \[ y = 4^x \] Swap the roles of x and y to begin finding the inverse function: \[ x = 4^y \] Now, solve for y by taking the logarithm with base 4 of both sides (since 4 is the base of the exponential function): \[ \log_4(x) = \log_4(4^y) \] By the properties of logarithms, \( \log_b(b^a) = a \), so: \[ \log_4(x) = y \] Therefore, the inverse function is: \[ y = \log_4(x) \] Looking at the options provided: A. \( y = x^{-4} \) - Incorrect, because this represents a power function, not a logarithmic function. B. \( y = (\frac{1}{4})^x \) - Incorrect, this is another exponential function, not the inverse of \( 4^x \). C. \( y = \log_4(x) \) - Correct, as proved above. D. \( y = - \log_4(x) \) - Incorrect, this is the negative of the logarithmic function. The correct answer is C: \( y = \log_4(x) \).
Finding the Value of y with Given x
To find the value of \( y \) using the known value of \( x \), which is \( -\frac{1}{2} \), we should substitute \( x \) into one of the original equations. Let's use the second equation for this example: \( 3x + y = 2 \) Now we substitute \( x \) with \( -\frac{1}{2} \): \( 3(-\frac{1}{2}) + y = 2 \) Solve for \( y \): \( -\frac{3}{2} + y = 2 \) \( y = 2 + \frac{3}{2} \) \( y = \frac{4}{2} + \frac{3}{2} \) \( y = \frac{7}{2} \) So the value of \( y \) is \( \frac{7}{2} \). Therefore, the correct approach is to use the second option (B), "Input the value of \( x \) into either of the original equations."
Solving for Length of Hypotenuse in Right-Angled Triangle
You have presented a right-angled triangle ABC with an altitude BD from the right angle B to the hypotenuse AC, creating two smaller right-angled triangles, ABD and BDC. You have provided several pieces of information: 1. The angle \( \angle ABC = 15^\circ \) 2. The angle \( \angle BAD = 45^\circ \) 3. The length of BD is equal to that of DC, i.e., \( BD = DC \) You've asked to solve for the value of y, where y is the length of AC. But there's no direct indication of how the lengths on the triangle relate to y. The given angles and conditions suggest that we can calculate the length of AC using trigonometric ratios. Since triangle ABD and BDC are both isosceles and right-angled, we know several relationships: 1. \( AB = BD \) (since \( \angle BAD = 45^\circ \) it is an isosceles right-angled triangle, so the two non-hypotenuse sides are equal) 2. \( BD = DC \) (given) 3. Therefore, \( AB = BD = BC \), and those lengths are all equal to each other. To find y, we need to use these relationships and the fact that \( \angle ABC = 15^\circ \). We'll use the trigonometric identity for the sine of a 15-degree angle, as we're given \( \angle ABC \) and we have a right-angled triangle. In right-angled triangle ABC: \[ \sin(15^\circ) = \frac{AB}{AC} = \frac{BD}{AC} \] We know that \( \sin(15^\circ) \) can be expressed using the half-angle identity: \[ \sin(15^\circ) = \sin(30^\circ / 2) = \sqrt{\frac{1 - \cos(30^\circ)}{2}} = \sqrt{\frac{1 - \sqrt{3}/2}{2}} \] Simplifying the expression under the square root gives us: \[ \sin(15^\circ) = \sqrt{\frac{2 - \sqrt{3}}{4}} \] Using this identity, we can relate the length of AC to BD: \[ \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{BD}{AC} \] Since \( BD = BC \) and \( BC = AB \), let's denote the length of BD (also AB and BC) as x. Then the above equation becomes: \[ \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{x}{y} \] We can solve for y by multiplying both sides by y and dividing by \( \sqrt{\frac{2 - \sqrt{3}}{4}} \): \[ y = \frac{x}{\sqrt{\frac{2 - \sqrt{3}}{4}}} \] However, we have not been provided the actual numerical length for x (BD, AB, or BC), so we cannot provide a numerical solution for y without additional information. If you have a length for BD, AB, or BC, you can substitute that value into the equation to solve for y.
Solving Linear Equations
The given equation is \( -3x - 3y = 3 \). To solve for one variable in terms of the other, you can isolate one variable on one side. Let's solve for \( y \) in terms of \( x \): First, add \( 3x \) to both sides to get: \[ -3y = 3x + 3 \] Next, divide each term by \( -3 \) to solve for \( y \): \[ y = -x - 1 \] Now, \( y \) is expressed in terms of \( x \). If you want to solve for \( x \) in terms of \( y \) instead, you would do a similar process but isolate \( x \) instead of \( y \).
Solving for y with Given Data Points
The image contains two parts of information: 1. \( y = 21 \) when \( x = 9 \) 2. \( y = ? \) when \( x = -6 \) To solve the second part of the question, we need to find a relationship between x and y that is consistent with the given data. However, with only one data point given (\( y = 21 \) when \( x = 9 \)), we cannot determine the exact relationship between x and y because there are infinitely many mathematical functions or relationships that could satisfy this single condition. To proceed, we need either a specific function relating x to y, or more data points that would allow us to infer such a relationship. Without this information, we cannot determine the value of y when \( x = -6 \). If you have additional information such as a mathematical equation that relates x to y, please provide it so that the question can be solved.
Finding the Inverse of a Function
To find the inverse of the function \( f(x) = \sqrt{x} - 2 \), we need to switch the roles of x and y and then solve for y. Here are the steps: 1. Write the original function with y: \( y = \sqrt{x} - 2 \). 2. Swap x and y: \( x = \sqrt{y} - 2 \). 3. Solve for y: Starting with the equation from step 2, we will isolate y: \[ x = \sqrt{y} - 2 \] \[ x + 2 = \sqrt{y} \] (Add 2 to both sides) Now we need to get rid of the square root by squaring both sides of the equation: \[ (x + 2)^2 = y \] So the inverse function \( f^{-1}(x) \) is: \[ f^{-1}(x) = (x + 2)^2 \]
Converting Equations to Slope-Intercept Form
The instructions in the image ask to write the equations in "k-form" by solving for y. "K-form" usually refers to the slope-intercept form of a linear equation, which is: y = kx + m where k is the slope and m is the y-intercept. In this context, we need to express each equation with y isolated on one side. Let's do this for each equation: a) y - 5x = 3 To solve for y, simply add 5x to both sides: y = 5x + 3 b) 7x + y + 4 = 0 To solve for y, subtract 7x and 4 from both sides: y = -7x - 4 c) 5y = 15x + 10 To solve for y, divide all terms by 5: y = 3x + 2 d) 2x - y = 9 To solve for y, add y to both sides and then subtract 9 from both sides: 2x = y + 9 y = 2x - 9 Now each equation is in the slope-intercept form, with y isolated on one side.
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