Tutor profile: Lily H.
Subject: Linear Algebra
Find the slope of the equation 15x-3y=9
Step 1: When finding the slope of a linear equation we need to rearrange the equation to make y the subject. 15x-3y=9 Therefore subtract 15x from both sides of the equation. -3y= -15x+9 Then divide all terms by -3 to make y the subject. y=5x-3 Therefore the slope of the equation is 5. When the equation is in the form y=mx+c the slope is m.
Find the equation of the tangent to the curve y = x^3 + 2x^2 at the point (1,6).
Step 1: The first step to solving for any tangent line is to find the slope. Since we have the equation, we can solve for the derivative. The derivative will give us the instantaneous slope at x=1. The derivative of the equation is: f'(x)= 3x^2+4x Step 2: Plug in x=1 into the derivative to find the instantaneous slope at the coordinate x=1. This step will tell us the gradient of the tangent equation. f'(x)= 3x^2+4x f'(1)= 3(1)^2+4(1) f'(x)=7 Thus, the slope is equal to 7. Step 3: Use the point-slope form to plug in what we have solved for and are given so far. (m= slope=7) and the point given is (1,6). y-y1=m(x-x1) y-6=7(x-1) Firstly expand brackets and then rearrange the equation to make y the subject. y-6=7x-7 y=7x-1 Therefore the final answer is y=7x-1 The slope of the tangent is 7 and the y-intercept of the tangent is -1.
Step 1: When attempting to solve for "x", you must first multiply out the factors. This means we expand the brackets. 10x+ 20 - x + 5 = 6x - 24 + 10 Step 2: The next step is to combine like terms. 9x + 25 = 6x -14 Step 3: Gather like terms on one side of the equation in order to solve for x. Remember, what you do on one side of the equation must be applied to the other side. We bring the 'x' terms to the left side of the equation. 9x + 25 = 6x -14 -6x -6x _______________ 3x+25= -14 -25 -25 _______________ 3x= -39 Step 4: Now, divide each side of the equation by 3 in order to solve for x. x= -39/3 Therefore x = -13
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