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Jack L.
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SAT
TutorMe
Question:

Find the error in the following sentence or select e if there is no error: (a) Looking through the (b) magnifying glass, the (c) centipede seemed (d) enormous to the girl. e) No error

Jack L.
Answer:

The correct answer is (c). The phrase "Looking through the magnifying glass" should modify girl as she is the one who is doing the looking (rather than the centipede). Therefore, a possible way to correct this sentence would be something like "Looking through the magnifying glass, the girl saw an enormous centipede". In the revised sentence, girl appears right after the modifying phrase; thus, this new sentence is correct.

ACT
TutorMe
Question:

A cube of side length 5 is inscribed within a sphere such that the longest diagonal of the cube (i.e. from one corner to its opposite corner) forms the diameter of the sphere. Find the volume inside the sphere but outside the cube: a) 125 b)$$ 125(\dfrac {1-\pi \sqrt{3}}{2} )$$ c) $$125(\dfrac {\pi \sqrt{3}}{2} -1)$$ d) $$\dfrac{125 \sqrt{3}}{2} \pi$$ e) The correct answer is not listed

Jack L.
Answer:

The answer is c) $$125(\dfrac {\pi \sqrt{3}}{2} -1)$$ We begin by finding the length of the diagonal of one face of the cube. We do this by using the Pythagorean theorem $$5^2 + 5^2=c^2$, so $c = \sqrt{50} = 5 \sqrt{2}$$. Now, we can set up another right triangle with the diagonal of one face of the cube, a side of the cube, and the cube's longest diagonal (which we label $c_2$). This will result in $$(5 \sqrt{2})^2 + 5^2 =c_2^2$$. Therefore, $$c_2 = \sqrt{75} = 5 \sqrt{3}$$. Thus, $$5 \sqrt{3}$$ is the sphere's diameter. This means that the radius is $$\dfrac{5}{2} \sqrt{3}$$. We can now plug this radius into the formula for the volume of a sphere $$V = \dfrac{4}{3} \pi r^3$$, so $$V=\dfrac{4}{3} \pi (\dfrac{5}{2} \sqrt{3})^3=\dfrac{4}{3} \pi \dfrac{375 \sqrt{3}}{8}= \dfrac{125 \sqrt{3}}{2} \pi$$. Now, we have the volume of the sphere we just need to deduct the volume of the cube which is equal to $$5^3=125$$. So, the volume inside the sphere and outside the cube is equal to $$125(\dfrac {\pi \sqrt{3}}{2} -1)$$. This is equal to answer choice c.

Algebra
TutorMe
Question:

Solve the quadratic equation: $$x^2+2x+7=0$$

Jack L.
Answer:

We proceed by using the quadratic formula that for an equation of the form $ax^2+bx+c=0$: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ In this case, we see that $b=2$, $a=1$, and $c=7$. Therefore, we can plug into the quadratic formula: $$x=\dfrac{-2 \pm \sqrt{2^2-4*1*7}}{2}$$ $$x=\dfrac{-2 \pm \sqrt{-24}}{2}$$ $$x=\dfrac{-2 \pm 2i \sqrt{6}}{2}$$ $$x= -1 \pm i \sqrt{6}$$ Thus, we have found our two solutions for the value of x.

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