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Tutor profile: Lucas M.

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Lucas M.
College math teacher for four years. Data Analyst for j2 Global.
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Questions

Subject:Set Theory

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Question:

Let $$k\in \mathbb{Z}$$. Take $$A = \{ n\in \mathbb{Z}: \; n|k\}$$ and $$B = \{ n\in \mathbb{Z}: \; n|k^2\}$$. Prove $$A\subseteq B$$.

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Lucas M.

Let $$x$$ be any element of $$A$$. We will show that $$x$$ must also be an element of $$B$$. Since $$x\in A$$ we have that $$x|k$$. This means $( \exists c \in \mathbb{Z} \;\;\; s.t. \;\;\; xc = k$) Now we have $$xc = k$$. If we square both sides of this equation we get $((xc)^2 = k^2 \;\;\;\;\;\;\;\mbox{ (1)}$)$( x^2c^2 = k^2 \;\;\;\;\;\;\;\;\mbox{ (2)}$)$(xxc^2 = k^2 \;\;\;\;\;\;\;\;\mbox{ (3)}$)$( x(xc^2) = k^2 \;\;\;\;\;\mbox{ (4)}$) Since $$x,c \in \mathbb{Z}$$ we have that $$xc^2 \in\mathbb{Z}$$. Hence $$xc^2 =m$$ for some $$m\in\mathbb{Z}$$. Then by (4) we get $( xm = k^2$) for some $$m\in \mathbb{Z}$$. It follows that $$x|k^2$$. By definition of $$B$$ we have that $$x\in B$$. Since $$x$$ was an arbitrary element of $$A$$ this must be true for all elements of $$A$$. Hence $$\forall x \in A, \; x\in B$$. Therefore $$A\subseteq B$$. $$\square$$

Subject:Calculus

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Question:

A man holds a rock over the side of a building and tosses it directly upward with a velocity of $$15$$m/s. If the the building is $$45$$m tall, how long will it take the rock to hit the ground? (Recall that acceleration due to gravity is $$-9.8m/s^2$$).

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Lucas M.

We know that acceleration is constant at $$-9.8m/s^2$$. So at any time $$t$$, the acceleration of the rock is given by $( a(t) = -9.8 \;\;\;\;\; \mbox{for } t\geq 0$) The indefinite integral of the acceleration function will give us the velocity function $$v(t)$$ for the rock. Integrating both sides with respect to $$t$$ we get $( v(t) = -9.8t + v_0 \;\;\;\;\; \mbox{for } t\geq 0$) where $$v_0$$ is the initial velocity of the rock. We know the initial velocity of the rock to be $$15m/s$$ so $$v_0 = 15$$ and we have $( v(t) = -9.8t + 15 \;\;\;\;\; \mbox{for } t\geq 0$) Hence the velocity of the rock at time $$t$$ is given by $$v(t)$$. The indefinite integral of the velocity function will give us the position function of the rock. Integrating both sides with respect to $$t$$ we get $( s(t) = \frac{-9.8}{2}t^2+15t+s_0 \;\;\;\;\; \mbox{for } t\geq 0$)$( s(t) = -4.9 t^2+15t+s_0 \;\;\;\;\; \mbox{for } t\geq 0$) where $$s_0$$ is the initial position. We know the initial position of the rock to be $$45m$$ so $$s_0 = 45$$ and we get $( s(t) = -4.9t^2+15t+45 \;\;\;\;\; \mbox{for } t\geq 0$) Hence the position of the rock at time $$t$$ is given by $$s(t)$$. We see at time $$t=0$$, the position of the rock is $$s(0) = 45$$ which is consistent with the information from the question. To answer when the rock hits the ground we want to know when $$s(t) = 0$$. We need to solve $( 4.9t^2+15t+45 = 0$) Using the quadratic formula we get two solutions $( t \simeq -1.86 \;\;\;\;\mbox{ and }\;\;\;\; t \simeq 4.93$) The position function describes the rock for $$t\geq 0$$ so we take the positive solution. Therefore the rock hits the ground after about $$4.93$$ seconds.

Subject:Algebra

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Question:

Determine the $$x$$ and $$y$$ intercepts for the graph of the function $$y=f(x)$$ where $$f(x) = 2x^2+9x-5$$. Leave your answers as ordered pairs (eg:$$(-1,5)$$).

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Lucas M.

We start with the $$y$$-intercept: The $$y$$-intercept of a graph occurs when $$x=0$$. Hence, to find the $$y$$-intercept of the graph of $$y=f(x)$$ we set $$x=0$$ and determine the value of $$y$$. $$x=0$$: Here we plug $$0$$ into the function $$f(x)$$ and get $(y = f(0) = 2(0)^2+9(0)-5 = -5$) So when $$x=0$$, $$y=-5$$ and we get the $$y$$-intercept is $$(0,-5)$$. $$x$$-intercept(s): To find the $$x$$-intercepts of $$y=f(x)$$ we need to know when $$f(x) = 0$$ (or when $$y=0$$. This is where the graph $$y=f(x)$$ will cross the $$x$$-axis. To get these values we set $$f(x)$$ equal to $$0$$ and solve for $$x$$. Hence, we need to solve $(2x^2+9x-5=0$) This is a quadratic equation so we know there may be one solution, two solutions, or no solutions. To determine the number of solutions we could use the discrimination ($$b^2 -4ac$$) but that will not be covered here. When solving a quadratic equation we may try to factor or use the quadratic formula. Let's use the factoring technique here. The leading coefficient of the quadratic expression is $$2$$ which adds a few more steps to factoring. First multiply $$2$$ and $$-5$$ together (find the product $$a\cdot c$$). We get $$-10$$. Then we want to find factors of $$-10$$ that add to $$9$$, the coefficient of $$x$$. Factors of $$-10$$ are $(-1,10 \;\;\;\;\; -2,5 \;\;\;\;\; 1,-10\;\;\;\;\; 2, -5$) We see that $$-1$$ and $$10$$ add to $$9$$. Use these numbers to break up the middle term of the quadratic expression $(2x^2+9x-5=0$)$(2x^2-x+10x-5=0$) From here, factor by grouping $(2x^2-x+10x-5=0$)$(x(2x-1)+5(2x-1)=0$)$((x+5)(2x-1)=0$) Now we can use the zero product property to get two solutions for $$x$$ $(x+5=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2x-1 = 0$)$(x=-5\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x=1/2$) It is good to check that these solution are correct. That is we want to check that $$f(-5) = 0$$ and $$f(1/2) = 0$$. Check: $( f(-5) = 2(-5)^2 + 9(-5) -5 = 50-45-5 = 0$)$( f(1/2) = 2(1/2)^2 + 9(1/2) -5 = 1/2+9/2-5 = 5-5= 0$) We have confirmed that $$f(-5) = 0$$ and $$f(-1/2) = 0$$. Thus the $$x$$-intercepts are $$(-5,0)$$ and $$(-1/2,0)$$. Final Answer: $$x$$-intercepts: $$(-5,0), \;\;(-1/2,0)$$ $$y$$-intercept: $$(0,-5)$$

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