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Angela R.
Tutor for 10+ Years, Math Teacher for 8 years in Public Education
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Geometry
TutorMe
Question:

What type of triangle has side lengths $$15$$cm, $$8$$ cm and $$9$$cm?

Angela R.
Answer:

Use the Pythagorean Theorem to classify the triangle. The classification is determined by the relationships below: If $$a^2+b^2>c^2$$ then the triangle is acute. If $$a^2+b^2=c^2$$ then the triangle is right. If $$a^2+b^2<c^2$$ then the triangle is obtuse. I remember it by the relationship side $$c$$ has to the other sides. If $$c^2$$ is smaller, then I think of "small" angles, which makes me think acute. If $$c^2$$ is bigger, then I think of "big" angles, which makes me think obtuse. ALWAYS use $$c$$ to represent the longest side. The other two lengths can be either $$a$$ or $$b$$. I always use $$a$$ to be the shortest side for consistency. Let, $$a=8, b=9, c=15$$. $$a^2+b^2\Box c^2$$ $$8^2+9^2 \Box 15^2$$ $$64+81 \Box 225$$ $$145\Box 225$$ What is the relationship between the two numbers? $$145<225$$ Because 145 is less than 225 ( $$c^2$$ is bigger) the triangle is Obtuse.

Statistics
TutorMe
Question:

Maria recently decided to get a better idea of how often she moved during her day. She tracked her steps using her Fitbit for two weeks. The data and 5 number summary are listed below. Use the $$1.5*IQR$$ Rule to determine if there are any outliers in her data. \begin{matrix} 8111 & 9591 & 11738 & 11006 & 7502 & 8325 & 8049 \\ 6938 & 12938 & 5184 & 9181 & 6115 & 9555 & 9718 \end{matrix} \begin{matrix} Min & Q_{1} & Med & Q_{3} & Max \\ 5184 & 7502 & 8753 & 9718 & 12938 \end{matrix}

Angela R.
Answer:

The $$1.5*IQR$$ Rule states that a data point is an outlier if it falls below $$(Q_{1}-1.5*IQR)$$ or above $$(Q_{3}+1.5*IQR)$$. $$IQR= Q_{3}-Q_{1}$$ We get this information from the given 5 number summary. Please let me know if you want to see how the 5 number summary is calculated with or without statistical software. $$IQR= 9718 - 7502$$ $$IQR= 2216$$ First, lets find the high boundary to see if there are any high outliers. $$(Q_{3}+1.5*IQR)$$. $$9718 + 1.5 * 2216$$ $$9718 + 3324$$ $$13042$$ The maximum data point we have is 12348 which is not greater than 13042 so we can conclude that there are no high outliers. Now, lets find the low boundary to see if there are any low outliers. $$(Q_{1}-1.5*IQR)$$. $$ 7502 - 1.5 * 2216$$ $$7502 - 3324$$ $$4178$$ The minimum data point we have is 5184 which is not less than 4178 so we can conclude that there are no low outliers. There are NO outliers in this data set.

Algebra
TutorMe
Question:

Sam throws a ball off of a building. The ball's height can be represented by the function $$f(x)=-x^2+10x+9$$ where $$x$$ is time in seconds. How long does it take for the ball to hit the ground?

Angela R.
Answer:

We are looking for when the ball hits the ground. That would be where the function intersections the $$x$$ axis. When a function crosses the $$x$$ axis, the value of $$f(x)=0$$. $$0=-x^2+10x+9$$ It is easiest to solve a quadratic function when the leading coefficient is positive, so we will multiply everything by $$-1$$. $$0=x^2-10x-9$$ There are several ways to solve a quadratic function from here. I will use the quadratic formula, but please let me know if you'd like to see a different method. $$0=x^2-10x-9$$ let $$a=1$$ $$b=-10$$ $$c=-9$$ $$x=\frac{10\pm \sqrt{(-10)^2-4*1*(-9)}}{2*1}$$ $$x=\frac{10\pm \sqrt{100+36}}{2}$$ $$x=\frac{10\pm \sqrt{136}}{2}$$ $$x=\frac{10\pm 2\sqrt{34}}{2}$$ $$x=5\pm \sqrt{34}$$ Because this is a real life problem, it does not make sense to the use the negative solution, so our answer would be: The ball hits the ground after $$5+ \sqrt{34}$$ seconds.

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