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Marc C.
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Geometry
TutorMe
Question:

Line $$l$$ goes through point $$a$$ (-1,4) and is perpendicular to the line y = 2x+1. Find the equation to line $$l$$.

Marc C.

Since y = 2x + 1, the slope of that line is equal to +2. To find a perpendicular line, the slope of the new line multiplied by +2 should equal -1. Therefore: Slope x 2 = -1 Slope = $$\frac{-1}{2}$$ Using the standard point-slope formula: y = mx + b m is determined previously as $$\frac{-1}{2}$$ y = $$\frac{-1}{2}$$x + b Using the point a, the x and y-value help us solve for b 4 = $$\frac{-1}{2}$$(-1) + b 4 = $$\frac{1}{2}$$ + b b = $$\frac{7}{2}$$ Therefore the equation of the line would be: y = $$\frac{-1}{2}$$x +$$\frac{7}{2}$$

Pre-Algebra
TutorMe
Question:

Solve the following: (6 - 2 x $$3^{2}$$)$$^{2}$$

Marc C.

1. Using the PEMDAS method, the value within the Parentheses must be determined first. (6 - 2 x 3$$^{2}$$)$$^{2}$$ WITHIN the parentheses, the next step is to solve for the exponent. 3$$^{2}$$ = 9 so: (6 - 2 x 9)$$^{2}$$ Next the multiplication step. 2 x 9 = 18 (6 - 18)$$^{2}$$ Subtraction follows leaving the final answer in the parentheses to equal: (-12)$$^{2}$$ 2. Restart PEMDAS for the rest of the equation. The only remaining part is the exponent. (-12)$$^{2}$$ = 144

Algebra
TutorMe
Question:

Given the following equation: f(x) = 5 - 2$$^{x}$$ Solve for f $$^{-1}$$(-3)

Marc C.

Solving for f $${^-1}$$(-3) indicates that the x-value of -3 needs to be inserted into the inverse of the equation f(x). In order to do so we begin with the function: y = 5-2$$^{x}$$ 1. We switch x and y x = 5-2$$^{y}$$ 2. Solve for y. Begin by subtracting -2$$^{y}$$ and x from both sides. 2$$^{y}$$ = 5-x Next, using log rules we can solve for y y = log$$_{2}$$(5-x) 3. Plug in -3 y = log$$_{2}$$(2$$^{3}$$) = 3 y = 3 f $$^{-1}$$(-3) = 3

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