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This means that 35 pastries were croissants, so the percentage is 35/100 = 35%.

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Can you describe the symbol more in depth? Is it a letter denoting a function, or a Greek letter, or absolute value, or a matrix, etc...

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The first pair of equations is quite simple, just add them to get

13y = 39, y = 3 --> x = 4

Do the others the same way. You might have to multiply one or both equations by a constant to cancel a variable out. For example, the next set of equations, you could multiply the first equation by 3 and the second equation by 2, then adding them to cancel y.

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(25 mi)*($.99/mi) = $24.75

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Reorder the terms to produce . By the quadratic formula,

. Since the roots have an imaginary part not equal to zero, the zeros of the function are complex.

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If you recall the right triangle with sides x, , and 2x, then if x = 8, the missing side is , or 13.856.

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t+3, t-3

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There's quite a bit to teach, as pre-algebra, algebra, and algebra (as a whole) are very broad subjects. I suggest you read your pre-algebra book or go to a library and find a pre- or algebra book.

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Replace y with 3x in the equation x = 3y to get

x = 3(3x) = 9x

x = 0

Since y = 3x = 3(0), y = 0.

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I would advise not to put your entire homework into one question. For nearly all the problems, you can solve them by using the identities and (regardless of the base). A couple of the problems will require you to use .

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The quadratic formula says that the two roots of any second degree polynomial {{ax^2 + bx + c}}} are .

It is possible to derive the quadratic formula by completing the square. This Wikipedia article lists multiple ways to derive the formula:

http://en.wikipedia.org/wiki/Quadratic_equation

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The equation you have becomes -2 = 40, by cancelling the x^2.

Instead, it should look something like

By difference of two squares this is equivalent to





, therefore the two integers are 11 and 9. It is easy to check that

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-5x + 2y is the simplest form.

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If x is the length, then the width y must be equal to 2x. So, y = 2x and we can substitute y to obtain

x + x + 2x + 2x = 18

6x = 18

x = 3 --> y = 2(3) = 6.

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Look at a unit circle. For any point on the unit circle, the x-coordinate represents and the y-coordinate represents .

Since cos x = -1, the x-coordinate must be -1, and the y-coordinate must be 0 (due to the Pythagorean identity). This only occurs when .

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If a linear equation is vertical, then its slope is undefined. The only way for (the expression for slope) to be undefined is if there is some such that . This means that for some change in y, x is held constant. Since the line goes through (-4, 8), then x is held constant at -4. Therefore the equation is x = -4.

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Yes. Except, use an alternate form, distance = rate*time.

D = (5 mi/hr)(.5 hr) = 2.5 mi

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Since , --> y = -1

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We want to show that . Rewriting everything in terms of sin and cos,

Cancel out sin(x) to obtain

which is true for all x that makes the values defined.

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It simply means that f(x) defines a function for a value x. It's similar to an input-output method. The function f(x) is linear, so the graph looks like this:

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I don't know what the original equation or point is.

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-5x

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You forgot the x terms in the expansion. It should look like:



After that, you can collect like terms and solve via the quadratic formula.

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Many different ways to solve this.
Solution 1
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If the sides of the rectangle are x and 50-x (so that the perimeter is 100), the area is , which is a parabola in terms of x. The vertex occurs at , or x = 25. It points downward, so x = 25 maximizes the area, so the area is 625.

Solution 2
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Another way is finding the extrema of . We have which is equal to zero when x = 25 (this is actually where the -b/2a rule comes from).
Solution 3
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Yet another way comes from an unusual theorem: AM-GM inequality. Suppose we have two terms and . Then by AM-GM,

Since , ,





The AM-GM inequality says the equality occurs if and only if all the 's are equal, that is, , and the optimal area is 625.


It's pretty rare you'll see a solution like the last one. However AM-GM can be used to prove many similar theorems, i.e. proving that the rectangular solid of fixed surface area that has maximum volume is a cube.

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Add the first two equations (this clears out the y-term) to get

4x + 2z = 20 --> 2x + z = 10

Multiply the second equation by 2, so we can add the second and third equations and also clear the y-terms. This results in another equation in terms of x and z that is not equivalent to the one we already obtained.

4x - 2y + 6z = 26
3x + 2y - z = 17
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7x + 5z = 43

We can rewrite this as 5(2x + z) - 3x = 43 --> 5(10) - 3x = 43 --> x = 7/3, z = 16/3. Substituting x and z into any of the equations, we get y = 23/3. Therefore the solution (x,y,z) is (7/3, 23/3, 16/3).

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Each angle must measure 60 degrees. Therefore, 2y = 60 --> y = 30.

Now use the first angle 3x + y, replace y with 30. It is equal to 60, so x is 10.

The remaining angle, 2x+40, is already equal to 60 since if we replace x with 10, we get 60.

So, x = 10, y = 30.

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Let the blank number be x. Add 10+50 to get 60, so we have

60 = 80 - x. Add x to both sides to get 60 + x = 80, subtract 60 to get x = 20. The second problem is very similar to the first one.

In terms of explaining to a 7-year old...I had a math tutor from a university who, even on the most difficult olympiad problems, would replace some expression like with an apple or pear, and do the problem in terms of apple. It worked quite well for me. You might want to try that with your son. This is similar to an elementary algebra problem, but might be more visual and appealing than using "x."

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Since AD || BC, angle 1 = angle 3, so you can replace the "3" with a "1." Also, since angle 2 = angle 1 (already given), AB = BC, and since AB = CD and BC = AD (the parallel sides of a parallelogram are congruent), all four sides are congruent to each other and it's a rhombus.

I almost never use the two-column proof when writing solutions to proof problems...it helps you back up each statement with a reason, but virtually no mathematicians use the two-column proof when explaining a proof(e.g. in a paper).

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By Vieta's formulas, the sum of the roots of a quadratic is . In this case, the sum of the roots is -3/5.

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I'm pretty sure you can easily solve both systems by adding the two equations together (besides, the problem tells you how to solve it). However, you'll have to multiply one of the equations on the second problem by -1, so that you can add and the y's will cancel out.

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With the information supplied, I don't have enough information since I only know two pairs of sides are congruent (as opposed to SSS). But if the angle in between is a right angle, then the triangles must be congruent by HL. Note that HL is a special case of SAS congruency.