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Jonathan C.
High School Tutor for Three Years
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Algebra
TutorMe
Question:

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

Jonathan C.

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
TutorMe
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.

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
TutorMe
Question:

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

Jonathan C.

\$\$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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