Enable contrast version

# Tutor profile: Jacob M.

Inactive
Jacob M.
Math Tutor for 5 years with a passion for teaching
Tutor Satisfaction Guarantee

## Questions

### Subject:Calculus

TutorMe
Question:

A ladder 13 meters long rests on horizontal ground and leans against a vertical wall. The top of the ladder is being pulled up the wall at 0.1 meters per second. How fast is the foot of the ladder approaching the wall when the foot of the ladder is 5 m from the wall?

Inactive
Jacob M.

This is a related rates problem. In order to solve this, we need to find an equation that relates how far up the wall the ladder is with how close the foot of the latter is to the wall, then use that equation to relate the rate of change of these variables to each other. Let's call the height of the latter $$y$$ and the distance of the foot to the wall $$x$$. Notice that the latter forms a triangle. This means we can use the Pythagorean theorem to relate these variables. Since the latter, which is length 13 meters, is the hypotenuse, that gives use the equation $$x^2+y^2=13^2$$. Since we want the rate of change of these variables over time, we need to differentiate this equation with respect to time, $$t$$. This gives us $$2x\frac{dx}{dt} + 2y\frac{dy}{dt}=0$$. This is using the power rule and also implicit differentiation, since this is done with respect to $$t$$. We want to know how fast the foot of the ladder is approaching the wall, which is $$\frac{dx}{dt}$$ so we solve for that to get $$\frac{dx}{dt} = -\frac{y}{x}\frac{dy}{dt}$$. We are told it is being pulled up at 0.1 m/s so $$\frac{dy}{dx}=0.1$$ and the foot of the latter is 5 m from the wall so $$x=5$$. Using this and the Pythagorean theorem, we get that $$y=12$$. Plug all of this in to get $$\frac{dx}{dt} = -\frac{12}{5}*0.1 = -\frac{6}{25}$$. Thus the foot of the ladder is moving toward the wall at a rate of $$\frac{6}{25}$$ m/s.

### Subject:Statistics

TutorMe
Question:

The probability that a weighted die lands on an even number is 80%. The die is rolled 4 times. Find the probability that all 4 are odd numbers.

Inactive
Jacob M.

First we need to find the probability of rolling an odd number. Since the probability of rolling an even number is 80%, by the complement rule, the probability of rolling an odd number is 1-0.8 = 0.2 or 20 percent. Since each roll is independent, we simply multiply the 4 probabilities together 0.2*0.2*0.2*0.2 = 0.0016 or 0.16%.

### Subject:Algebra

TutorMe
Question:

Factor the polynomial $$3x^7+6x^5-9x^3$$.

Inactive
Jacob M.

First we need to factor out the greatest common factor. Since each term is divisible by 3 and each $$x$$ is raised at least to the third power, the greatest common factor is $$3x^3$$. Now we have $$3x^3(x^4+2x^2-3)$$. To factor $$x^4+2x^2-3$$, we need to turn this into a quadratic. If we say $$X=x^2$$ we can rewrite this part as $$X^2+2X-3$$. Now to factor this we use AC method. We must find two numbers that multiply to $$-3$$ and add to $$2$$. Those two numbers are $$3$$ and $$-1$$ which means this factors as $$(X+3)(X-1)$$. Substituting $x^2$ back for $X$ we have the expression is equal to $$3x^3(x^2+3)(x^2-1)$$. We are close, but now $$x^2-1$$ will factor further. Since this is a difference of squares, this factors to $$(x+1)(x-1)$$. So we have a final answer of $$3x^3(x^2+3)(x+1)(x-1)$$.

## Contact tutor

Send a message explaining your
needs and Jacob will reply soon.
Contact Jacob

Start Lesson

## FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.