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Jonathan C.

High School Tutor for Three Years

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Algebra

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Question:

Factor $$f(x)=x^4 -16$$

Jonathan C.

Answer:

DOTS (Difference of Two Squares) states that $$a^2-b^2 = (a-b)(a+b)$$ In this question, $$a^2 = x^4$$ $$b^2 = 16$$ Taking the square root of both sides gives: $$a=x^2$$ $$b=4$$ Therefore, $$(x^4-16) = (x^2-4)(x^2+4)$$ (Equation 1) However, $$(x^2-4)$$ can also be factored. Using DOTS again, $$a^2 = x^2$$ $$b^2 = 4$$ Taking the square root of both sides gives: $$a=x$$ $$b=2$$ Therefore, $$(x^2-4) = (x-2)(x+2)$$ Plugging this back into Equation 1, $$(x^4-16) = (x^2-4)(x^2+4) = (x-2)(x+2)(x^2+4)$$ $$(x^4-16) = (x-2)(x+2)(x^2+4)$$

Physics

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Question:

A ball is thrown horizontally with initial velocity 10.0 m/s from the top of a 10.0m cliff. What distance away from the base of the cliff does the ball land? (Neglect air resistance)

Jonathan C.

Answer:

First, we need to find the time that the ball travels before hitting the ground. Since there is constant downward acceleration and there is no vertical component to the ball's initial velocity, $$ a_y = -9.81\ m/s^2$$ and $$ v_{0y} = 0\ m/s$$. Therefore we can use the following kinematics formula. $$y = y_0 + v_0t + .5at^2$$ $$0\ m = 10.0\ m + (0\ m/s)(t) + .5(-9.81\ m/s^2)(t^2)$$ $$-10.0\ m = -4.905\ m/s^2(t^2)$$ $$t^2 = 2.04\ s^2$$ $$t = 1.43\ s$$ Since there are no other forces acting on the ball in the x direction after release, $$a_x = 0\ m/s^2$$ We can now use the following kinematics formula. $$x = x_0 + v_xt$$ $$x = 0\ m + (10\ m/s)(1.43\ s)$$ $$x = 14.3\ m$$ The ball will land 14.3 m away from the base of the cliff.

Macroeconomics

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Question:

If the marginal propensity to save is 40%, what is the spending multiplier?

Jonathan C.

Answer:

$$Spending\ Multiplier = 1/(Marginal\ Propensity\ to\ Save)$$ $$Marginal\ Propensity\ to\ Save = 40\% = .40$$ $$Spending\ Multiplier = 1/.40 = 2.5$$ The spending multiplier is 2.5

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