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Set up systems of equations:

We have a+c=60 and 8a+5c=420.

\(\begin{bmatrix} 1 & 1 \\ 8 & 5 \end{bmatrix}\)\(\begin{bmatrix} 60\\420 \end{bmatrix}\), and we take the determinant of the first matrix, which is \(D\), which can be found by solving \(ad-bc\), and that gives \((1)(5)-(1)(8)=5-8=-3\).

Now, we have to find the inverse of this matrix, \(\begin{bmatrix} 1 & 1 \\ 8 & 5 \end{bmatrix}\), where you have to switch the \(a\) and \(d\), and find the negative values of \(b\) and \(c\).

Thus, the new matrix is: \(\begin{bmatrix} 5 & -1 \\ -8 & 1 \end{bmatrix}\). And, now we have this: \(\frac{1}{-3}\begin{bmatrix} 5 & -1 \\ -8 & 1 \end{bmatrix}\), and "distributing" them \(-3\) to all the terms in the matrix, we get(have) \(\begin{bmatrix} \frac{5}{-3} & \frac{-1}{-3} \\ \frac{-8}{-3} & \frac{1}{-3} \end{bmatrix}\). and this can be re-written as \(\begin{bmatrix} \frac{-5}{3} & \frac{1}{3} \\ \frac{8}{3} & \frac{-1}{3} \end{bmatrix}\). Now, we have to multiply \(\begin{bmatrix} \frac{-5}{3} & \frac{1}{3} \\ \frac{8}{3} & \frac{-1}{3} \end{bmatrix} * \begin{bmatrix} 60\\420 \end{bmatrix}\), and multiplying the first column by the first row, and the first column by the second row! (The second matrix has only one column), we get \(\begin{bmatrix} 40\\20 \end{bmatrix}\). Thus, the answer is forty\((40)\) almonds and twenty\((20)\) cashews.

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Using the Principle of Inclusion and Exclusion, we have \(63+85+62-31-25-20+8=142\) students. Thus, eight of them like neither of the three subjects.

Math Only: 63-31-25+8=7+8=15 students.

English Only: 85-31-20+8=42 students.

History Only: 62-25-20+8=25 students.

We have n+d=70

5n+10d=555

Set up matrices:\(\begin{bmatrix} 1&1 \\ 5&10 \end{bmatrix}\)\(\begin{bmatrix} 70\\ 555 \end{bmatrix}\). Take the determinant of the first matrix, and that gives \(ad-bc=10-5=5.\) Finding the inverse of the matrix or \(A^{-1}\), we get \(\frac{1}{5}\begin{bmatrix} 10&-1 \\ -5&1 \end{bmatrix}\)\(\begin{bmatrix} 70\\ 555 \end{bmatrix}\), and dividing each term, gives and gets us \(\begin{bmatrix} 2 & \frac{-1}{5} \\ -1 & \frac{1}{5} \end{bmatrix}\)\(\begin{bmatrix} 70\\555 \end{bmatrix}\), and multiplying both of them together, take the first row of the first matrix and multiply it by the first column of the second matrix, then multiply the second row of the first matrix with the second column of the second matrix to get an answer of \(\begin{bmatrix} 29\\41 \end{bmatrix}.\) And testing the values gives us, twenty-nine(29) nickels and forty-one(41) dimes.

Systems of Equations are much easier!

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Let's see if I can attack both of the questions.

(a): The area of a regular hexagon can be found by splitting the hexagon into six equilateral triangles, using 30-60-90 triangles by drawing an altitude and multiply this results by six to get the area answer. However, a very popular formula for a hexagon is \(\frac{3\sqrt{3}}{2}s^2\) where \(s\) is the side length of the hexagon, and we will get an area(answer) of \(\boxed{216\sqrt{3}}\) centimeters squared.

-We can also solve this first part of the problem by noting the area of a regular figure or shape is half its perimeter times the distance from the center of the polygon to a side. (Try to prove this on your own!) And, the apothem is the distance from the center of a polygon to a side. We can generate and generalize the formula \(\frac{1}{2}*P*A\), where \(P\) is the perimeter and \(A\) is the apothem. Next, the area of the hexagon is \(12*6=72cm\), because the side length of the hexagon is twelve centimeters and a regular hexagon has six equal sides. We have the apothem as \(6\sqrt{3}\) since given in the problem. Thus,l the area is \(\frac{1}{2}*72*6\sqrt{3}=\boxed{216\sqrt{3}}\) centimeters squared. 

(b): The perimeter is quite easier than calculating the area since the side length that is given in the problem is twelve centimeters, and a regular hexagon has six equal sides. Thus, the perimeter is \(12*6=72\)centimeters.

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We have the sides x units, 33 units, and 42 units.

Thus, we have x>42-33

x>9

x=10 units.

Thus, the smallest possible perimeter will be \(10+33+42=43+42=\boxed{85}\) units.

The total number of students: 10 C6=210.

Boys: 6C3 and Girls: 4C3=4, 20*4=80.'

Thus, the answer is 80/210=8/21. 

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(x+y)^2=x^2+y^2+2xy=78

x^2+y^2+2(13)=78

x^2+y^2=52

(x-y)^2=x^2+y^2-2xy, 52-2(13)=52-26=26,

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Every number appears Pascals' Triangle at least once, so the answer is a hundred(100).

I might as well post my solution.

We can use a very handy formula, which is very important to remember, \(A=rs\), where \(A\) is the area of the figure, \(r\) is the inradius when it is a complete circle, and \(s\), the semi-perimeter of the figure, or you can call it as, "half the perimeter."

Now, this is where the tricky part comes in. we can reflect the second part of the figure, like this:

After drawing this figure, we can find the area of one triangle, the multiply it by two, to find the area of the rhombus.  Drawing the altitude of sixteen to the base, we find that the area is \(\frac{1}{2}*16*15=120\). Multiply it by two since there are two triangles, to get \(2*120=240.\)And, using the Pythagorean Theorem, that gives us \(\sqrt{8^2+15^2}=\sqrt{17^2}=17.\) Multiply it by four to find the whole(full) perimeter, and divide it by two to find the semi-perimeter, to get \(17*2=34.\) Now, we found \(A\) and \(s\), and we divide it to get \(r\), to get \(\frac{240}{34}=\boxed{\frac{120}{17}}.\)

Partial Fraction Decomposition does the job.