Math Test 2 3.4-6.1
(4.1) Solve the system by graphing x + y = -5 -2x + y = 1
(-2,-3)
(4.1) Solve the system by elimination. 1/5x + 1/7y = 12/5 1/10x + 1/3y = 5/6
(7)1/5x+1/7(5)=12/5(7)->7/35x+5/35y=84/35 -> 7x+5y=84 (-2) -> -14x-10y=-168 3x+10y=25 -> 3x+10y=25 ->-11x=-143-> x=-13 7(-13)+5y=84-> 5y=7 -> y=7/5 ANSWER: (-13, 7/5)
(3.6) For the function find (a) f(2) and (b) f(-1) (EQUATION:) f={(-2,2),(-1,-1),(2,-1)}
(a) -1, (b) -1
(4.1) Solve the system by elimination. -2x + 3y = -16 2x - 5y = 24
-2x + 3y = -16 -> 2x - 5y = 24 -> 0 - 2y = 8 -> y = - 4 -2x + 3(-4) = -16 -> -2x - 12 = -16 -> -2x = -4 -> x = 2 ANSWER: (2, -4)
(4.1) Solve the system by substitution. 1/2x + 1/3y = 3 -3x + y = 0
1/2x+1/3(3/1)=3 -> 1/2x+1=3 -> 1/2x=2 -> 1/2 * 2/1 = 2/1 -> x = 2 y=3x+0 -> y=3(2)+0 -> y=6 ANSWER: (2, 6)
(4.3) Venus and Serena measured a tennis court and found that it was 42 ft. longer than it was wide and had a perimeter of 228 ft. what was the length and the width of the tennis court?
2L+2W=228 L=W+42 2(W+42)+2W=228 -> 2W+84+2W=228 -> 4W+84=228 -> 4W=144 -> W=36 L=36+42 -> L=78 CHECK: 78(2)+36(2)=228
(4.1) Solve the system by substitution. 4x + y = 6 y = 2x
4x+(2x)=6 -> 6x=6 -> x = 1 4(1) + y = 6 -> 4 + y = 6 -> y = 2 ANSWER: (1, 2)
(4.1) Write the equation in slope-intercept form and then tell how many solutions the system has. 3x+7y=4 6x+14y=3
7y=-3x+4 -> y=-3/7x+4/7 14y=-6x+3 -> y=-6/14x+3/14 -> y=-3/7x+3/14
(4.1) Decide whether the given ordered pair is a solution of the given system: (8, -9) x - y = 17 x + y = 1
8-(-9)=17 -> 8 + 9 = 17 (TRUE) 8 - 9 = 1 (TRUE)
(3.5) Decide whether the relation defines y as a function of x. (Solve y first if necessary.) Give the domain. y=(sqrt) x-7
All real numbers greater or equal to seven. (Because you can't square root negatives, you have to make sure when you subtract 7 you don't get a negative number) x≥7 OR [7,∞)
(3.5) Decide whether the relation defines y as a function of x. (Solve y first if necessary.) Give the domain. y=-(sqrt)x
All real numbers greater or equal to zero. (Because you can't square root a negative, so it has to be any real numbers that aren't negatives.) x≥0 OR [0, ∞)
(3.5) Decide whether the relation defines a function, and give the domain and range. {(5,1), (3,2), (4,9), (7,6)}
FUNCTION. Domain: (5, 3, 4, 7) RANGE: (1, 2, 9, 6)
(3.4) Graph the linear inequality in two variables x+4≤-3
Points: (-1, -2), (3,0)
(3.4) Graph the compound inequality 3x-y≥3 and y<3
Points: (0, -3), (3,6)
(3.4) Graph the linear inequality in two variables x+y≤2
Points: (0,2), (4,-2)
(3.6) Let f(x)=-3x+4 and g(x)=-x^2+4x+1 (EQUATION:) f(-3)
f(-3)=-3x+4=13 (x=-3 [plug in] -3(-3)+4 -> 9+4=13)
(3.6) Let f(x)=-3x+4 and g(x)=-x^2+4x+1 (EQUATION:) g(-2)
g(-2)=-(-2)^2+4(-2)+1= -4-8+1=-11
(3.5) Decide whether the relation defines a function, and give the domain and range. {(2,4), (0,2), (2,5)}
NOT A FUNCTION Domain: (2,0) Range: (4,2,5)