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## Answers by Anthony P.

This is a system of equations; two equations, two unknows. Translate the words into expressions and, ultimately, to equations. The area of Desert A is nine times the area of Desert B. A = 9B (eqn 1) the sum of their areas is 1,000,000 square miles A + B = 1,000,000 (eqn 2) Now,...

To graph, simply create a table of values for x and y (when x is __,  y is __ ). Although if you know the basic parent function of ln(x), then this one is simply shifted to the left by three units. For the domain, that means all values of x that satisfy the equation. Therefore, we have...

Is the value of x = 4/3? If so, then substitute the (4/3) into the expression and evaluate. 18(4/3) - 3 18 times (4/3) minus 3 Order of operations: parentheses, exponents, multiplication/division, addition/subtraction Multiplication/division before addition/subtraction. So...

I assume the problem is stated as: log2(x4 sqrt(y/x3)) Learn the 4 basic properties of logs: (1) logb(xy) = logbx + logby. (2) logb(x/y) = logbx - logby. (3) logb(xn) = n logbx. (4) logbx = logax / logab. (to simplify typing, I'll temporarily omit the base on...

A(t) = Aoe(kt) We first need to solve for the rate constant. When t = 1 min (i.e., A(1) ), we are told that 45% dies leaving 55% to remain (0.55 * 5x10^6 = 2.75x10^6 = A(1)) 2.75x10^6 = 5x10^6 e^(1k) 2.75x10^6 / 5x10^6 = e^k Take the natural log of each side: ln(2.75x10^6/5x10^6)...

This is a compound interest formula, where P = principal, r = rate, n = # compoundings per yr, t = time (in years). P = 3000 r = 7% = 0.07 t = 6 n = 2 (semi-annually) a) A = 3000(1 + 0.07/2)^(2*6) = \$4,533.21 b) A = 3000(1 + 0.07/12)^(12*6) = \$4,560.32  Continuous compounding:...

C*(0.08) = 1740 Divide each side by 0.08 C = 1740/0.08 = 21750

The domain are values of x that satisfy the equation, then the range would be the corresponding values of y at the x-values of -5, -2, 7. Therefore plug each value of x in the equation and evaluate to find the corresponding y-values. Those will be your range. e.g., y = -2/3(-5) + 4 =...

This is a "plug and chug" problem. The only snag would be to always make sure the units are S.I. The force of gravity between two objects is jointly proportional to the object's masses and inversely proportional to the square of the distance: F = G(M1*M2) / r2 F = 6.67E-11...

2(3x - 2) ≥ 22 Solve like a typical eqn Distribute. 6x - 4 ≥ 22 Add 4 to each side. 3x ≥ 26 Divide by 3 on each side x ≥ 26/3 or 8 2/3 A value of x that makes the inequality statement true is therefore any that is ≥ 9 1/3. (Just remember that...

(1) y = -3x + 5 (2) 5x - 4y = -3 Substitute eqn 1 into eqn 2 for y. (It doesn't matter which eqn or which variable, but eqn 1 was already solved for y.) 5x - 4(-3x + 5) = -3 Simplify, and solve for x. 5x +12x -20 = -3 17x - 20 = -3 17x = 17 x = 1 Now...

5 + 2(4x) - 2(4) = 3(2x) - 3(1) 5 + 8x - 8 = 6x - 3 8x - 3 = 6x - 3     +3         +3 8x = 6x - 0 -6x   -6x 2x = 0 x = 0

Solve for any variable from either equation: (1) 4x - y = 3  ==> y = 4x - 3 Substitute into other equation: (2) 9x - 2y = 8 9x - 2(4x - 3) = 8 Solve: 9x - 8x + 6 = 8 x + 6 = 8 x = 2 Substitute to find y: y = 4x - 3 = 4(2) - 3 y = 8 - 3...

P = 38 L = W + 5 P = 2L + 2W P = 2(W + 5) + 2W = 2W + 10 + 2W  38 = 4W + 10 28 = 4W W = 7 ft L = W + 5 = 7 + 5 = 12 ft

d - 2(d - 3) + 5d ≥ 36 Solve using steps similar to ordinary equations: d - 2d + 6 + 5d ≥ 36 4d + 6 ≥ 36 4d ≥ 30 d ≥ 30/4 d ≥ 15/2 So d must be a value that is greater than or equal to 15/2.       ...

r2 - 4r - 3 = 6 Normally, we can solve by factoring the equation (if factorable) and setting each factor equal to zero. First we need to get the equation in standard form (Ax2 + Bx + C = 0) by subtracting 6 from each side: r2 - 4r - 9 = 0 Now, we factor the trinomial into the...

Watts are a unit of power (unit energy per second in this case).  The utility company charges \$0.10 per kilowatt-hour, or 0.10/kW*hr. To do the conversion correctly, we need to convert watts to kilowatts. So 350 W = 0.350 kW. Over the course of one yr: \$0.10  x 24 hr x 365...

A trinomial, if factorable, will factor into the product of two binomials. The method to factor this one is fairly straight forward since the coefficient of the first term is 1. We are essentially "undoing" the result of FOIL'ing. (a + b)(a + b) = a2 + 2ab + b2 First...