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What model?

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Slope = 1

x-intercept occurs when y = 0, i.e. 0 = x+3, x = -3. x-intercept: (-3,0)

y-intercept occurs when x = 0, y = 0+3 = 3. y-intercept: (0,3)

Graph of the function:
http://www.wolframalpha.com/input/?i=y%3Dx%2B3

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x^2 + x - 56 = 0 when x = 7 or x = -8 (you can check by factoring). It can be shown that x^2 + x - 56 is positive when x > 7 or x < -8. Hence there is no smallest possible value of x (since x can go to negative infinity and the area is still positive). Note that the dimensions themselves do not necessarily have to be polynomials.

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Surely you don't expect tutors on algebra.com to own the exact same textbook that you are using. You should explain the problem more in depth.

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Yes; if you were to expand, you obtain f(x) = x^2 - x. Alternatively, you could claim that the product of two polynomials is a polynomial.

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You can let where k is an integer, then . Cubing both sides,



Here, we can take this equation "modulo 1" by eliminating all the integer expressions (b,c,k^3,3ka).



Factor LHS

Note that you can replace with . Modulo 1, this is equivalent to -sqrt(a).





Here, we show that i.e. it is an integer. I'll let you finish the proof that sqrt{a}, sqrt[3]{b}, and sqrt[3]{c} have to be integers. Pretty daunting problem...unfortunately we cannot assume the converse of the statement is true (i.e. if ... are integers then ... is an integer).

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Assuming (h,k) is the center of the circle, complete the square:





The coordinates of the center are (-8,2).

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Statements a) and b) cannot be true, since it is possible to find counterexamples for each one. Note that a cyclic quadrilateral is a quadrilateral whose four vertices all lie on a circle.

For statement a), a "kite" shape formed by taking a right triangle and reflecting it about the hypotenuse is a counterexample, since such a quadrilateral is cyclic and not every kite shape will have diagonals that bisect each other.

For statement b), a parallelogram (other than a rectangle) is a counterexample, because its diagonals bisect each other, but parallelograms (other than rectangles) *cannot* be cyclic (since opposite angles must add to 180).

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"Complementary," not "complimentary."

Let 11x and 5x be the angles. Since they're complementary, their sum is 90, so 11x+5x = 90. Solve for x, then find 11x and 5x.

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2-2 = 0, not much of an arithmetic property. However, the number -2 functions as the "inverse element" of 2. Generally speaking, if e is an identity element, * is a binary operation, and x is some number (2 in this case) then there exists a y such that x*y = e for all x.

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Note that



and that variance is equal to the square of the standard deviation. The best method is simply to use a calculator to input the data values, and you should obtain

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Assuming log is in base 10, your solution is correct. You want to make sure if the log is in base 10 or in base e (i.e. ln)

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Use A = (1/2)bh

69 = (1/2)(13)b

Solve for b

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is not equal to 1. You should ask yourself, "8 to what power equals 4?" It turns out that the logarithm is equal to 2/3.

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You'll need the cubic formula.

http://en.wikipedia.org/wiki/Cubic_formula#General_formula_of_roots

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For #1-3, review what natural numbers, rational numbers, integers, etc. are defined as.

For #4, x^0 = 1 because x^(a-b) = (x^a)(x^b), so if a = b, x^0 = (x^a)/(x^a) = 1 (assume x is not 0).

For #5, 5/0 is undefined because we cannot divide 5 into groups of zero.

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Suppose the coordinates are (0,0), (a,b), (c,d) and (a+c,b+d) (we may assume one of the coordinates is (0,0) in order to fix a point). Then the area is equal to the determinant of the matrix
|a b|
|c d|

or ad-bc.

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17 and 24, for example.

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1 furlong = 1/8 mile

4 furlongs = 4/8 = 1/2 mile

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Perimeter = 30+30+20+20 = 100 cm

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7^2 + 9^2 = 130

You could do it by letting x^2 + (x+2)^2 = 130, but since we're given that they're integers, we do not need to use the longer method.

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2 cos (n*pi) is not monotonic. Monotonic means that the function is either strictly increasing or strictly decreasing.

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There is no February 30...

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You could find the derivative of the function, i.e. velocity vs. time.

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9(x+y)

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1, 1/4, 1/9, 1/16, ... is not an arithmetic sequence.

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Cosine alternates between -1 and 1. cos x = 1 when x = -pi, -3pi, -5pi, etc.

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That is not an arithmetic sequence. Arithmetic sequences have consecutive terms that differ by a constant number.

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If a,b,c denote the digits, then

(right away we know the number is a multiple of 17)



Here, we conclude that a is even (since RHS is even). Now consider the equation mod 10 (if you don't know modular arithmetic, think of it as "matching the units digits on LHS and RHS").

If a = 2, then b = 4 or 9, but neither allows for an integer value of c. If a = 4, then b = 3 or 8, but likewise, there are no integer solutions for c. If a = 6, then b = 2 or 7, and it can be checked that b = 2 allows for the number 629. If a = 8, then b = 1 or 6, but neither allows for integer solutions.

Hence the only number is 629.

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Maria can run (how many) laps in the time it takes Susan to run three times around the track?

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We know what g(x) is, so g(x) + 5 is just "5 more than g(x)."