jfan17

I'm not trying to be rude, but did you take the time to read through CU's solution? It was an example problem, not the actual problem itself. What CU intends to do here is teach you the steps of the problem, not just how to read the answer and enter it in some website. Please take the time to actually try to learn what he's teaching you here. Using his method, you would get the answer of 83, which you already said yourself is wrong.

not if the question doesn't say to do so... for all we know the question wants the exact value in dollars

Ribeye steak = x 

Grilled salmon = y

On the first scenario:

12x + 30y = 557.59

In the second scenario:

21x + 10y = 582.75

If we multiply our second equation by 3, we get:

63x + 30y = 1748.25

We can then subtract our first equation from this new equation, to get:

51x = 1748.25-557.59 = 1190.66

x = $23.34627450980392

Plugging this into our first equation, we get:

280.1552941176470 + 30y = 557.59

30y = 277.434705882353

y = $9.24782352941177

The ribeye steak dinner is then $23.34627450980392, and the Grilled salmon is $9.24782352941177

Remember, dividing by a fraction is equivalent to multiplying by its reciprocal.

1/(1/3) = 1* (3/1) = 3. The answer is then 3, not 1/3.

You can think of it as dividing a stick of length one into lengths of 1/3 ; there are 3 of these smaller segments, not 1/3 of these smaller segments(that wouldn't make sense)

Geno how did you get the sum as close to 1/8? just asking

Any time!

1. Modulus is equivalent to saying magnitude. The magnitude of 2-3i is equal to:

\(\sqrt{2^2+(-3)^2} = \sqrt{4 +9} = \sqrt{13}\)

2. Multiplying 2-3i by 2 makes it 4 -6i. Doing the same thing as the first problem, the modulus is equivalent to:

\(\)\(\sqrt{4^2+(-6)^2} = \sqrt{16 +36} = \sqrt{52} = 2\sqrt13\)

3. Multiplying a complex number, a+bi, by a real number, c, will turn it into ac + bci. This is just a scalar of the original complex number. (it's the original multiplied by a factor)

Hi TGO, I'm not EP, but I'll give it my best shot.

Often times with problems like these, you start by looking at the discriminant of given quadratic. The discriminant is defined in an equation \(ax^2+bx+c = 0\) as \(\sqrt{b^2-4ac}\). If the discriminant is negative, then the roots are imaginary, and if it is non negative, then the roots are real. Let's look at the discriminant of this equation:

Given the quadratic

\(x^2+mx+4\)

The discriminant would be:

\(\sqrt{m^2-16}\)

We can write the inequality:

\(\sqrt{m^2-16} \geq 0\)

Squaring both sides, we get:

\(m^2-16 \geq 0\)

This factors out to:

\((m+4) * (m-4) \geq 0\)

With these types of problems, it's best to create a sign chart!

Although I don't have the means to create one as of now, I will explain all the cases in this "chart". 

The first case is where: 

\(m < -4\)

If this is true, then the first term(m+4) would be negative, and so would the second term, meaning that a negative * negative = positive, meaning that when m < -4, the inequality holds true.

The second case is where:

\(-4\leq m \leq 4\).

Testing numbers in this range, it's clear that none of them work. If m is negative but still greater than -4, the first term is positive, but the second negative, making the entire expression negative. If m is positive but less than 4, the second term is negative, and the first term is positive, still leaving the expression unsatisfied. It's clear that this interval doesn't work. Also realize that m can't be -4 or 4, because that would make the discriminant 0, meaning there's a double root(two of the same root). However, the problem says that the roots must be distinct

The third case is where:

\(4 < m\)

In this case, both terms are left positive, satisfying the inequality. 

This means that there are two intervals which we have:

\((-\infty,-4) \cup (4,\infty)\)

3. Let's first start off with the internal angle measurement of a regular hexagon. Each angle is equal to:

\(180-\frac{360}6 = 120\) degrees. If we drop a perpendicular to the smaller hexagon from a point like A, then we get two 30-60-90 triangles(since the smaller hexagon has interior angle measurement of 120, and the angles must add up to 180, so 120 + 30 + 30 = 180).

Let's then define the length of a side of the larger hexagon as x, meaning that the midpoint segments have length \(\frac{x}2\). 

First, let's find the area of the larger hexagon. Because a regular hexagon is comprised of 6 equilateral triangles(they all have the same side length, with 60 degree interior angle measurements), the area is equal to the area of one of the equilateral triangles multiplied by 6. Given a side length x of an equilateral triangle, the area is equal to:

\((x^2*\sqrt{3})/4\), which can be derived from heron's formula or pythagoras' theorem( just name an arbitrary side of the equilateral triangle as x). With this in mind, the area of a regular hexagon with side length x would then be:

\((6*x^2*\sqrt{3})/4 = (3*x^2*\sqrt3)/2\)

Now that we have the area of our larger hexagon in terms of x, we need to find the area of the smaller hexagon in terms of x. We can accomplish this by dropping an altitude from point A to the smaller regular hexagon, forming a perpendicular bisector with that side.

We then have two 30 - 60 -90 right triangles, with the hypotenuse having length \(\frac{x}2\), and the 60 degree angle being at the angle bisector of \(\angle FAB\). The 60 degree leg(the side adjacent to the smaller hexagon) is then \(x/2 * \sqrt3/3\) (remember the ratios of a 30-60-90 triangle!).

We then have that the entire side of the smaller hexagon is \(2*x/2 * \sqrt3/3\)(because two 30-60-90 triangles make up the side of the smaller hexagon). The smaller side is then \(x* \sqrt3/3\). Using our area formula from earlier, we have that the area is equal to:

\(3*(x*\sqrt3/3)^2*\sqrt3/4 = x^2*\sqrt3/4\)

Our fraction is then equal to:

\((x^2*\sqrt3/4) / ((3*x^2*\sqrt3)/2) = (x^2*\sqrt3)/4 * 2/(3x^2*\sqrt3)\)

Cross multiplying and canceling out terms, we get:

\(2/12 = 1/6\) As our answer.

This method of finding the fraction is certainly more extreme in my opinion than other much faster solutions. One such way is just assuming the side length of the regular hexagon because the fraction will stay constant no matter the side length.

2. Please reupload your question, the side lengths of the rhombus appear to be missing from the problem.