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  1. 1)45
    2)5
    3)20
    4)x-9=17
    5) 2x+3=35
    6)0
    7)-2/9(THIS IS NOT IN THE CHOICES BUT IT IS THE RIGHT ANSWER).
    8)x-(-2)
    9)Subtract 7 and then divide by -2
    10)-16
    11)5x=11
    12)Division property
    13)-27
    14)20
    15)2x-5=15
    16)-16
    17)168
    18)-5

  2. 1. X+9
    2. B-5+6
    3. 15+ 5 multiply V
    4. 2-3 multiply K
    5. 12 divide (H+2)
    6. M/7
    7. 2 multiply F + 9
    8. 6- (X-3)
    9. G/3 - 10
    10.4+ (4 multiply 5 multiply A)
    11. R-S + 14
    Hope this answer help you

  3. 1. x/-3=-15
    x=45

    2. -7+3(-12)÷-3
    -7+12
    5

    3. f(-4)=(-4)²-(-4)
    f(-4)=16-(-4)
    f(-4)=20

    4. x-9=17

    5. 2x+3=35

    6. -8-12-(-20)
    -20-(-20)
    0

    7. -2(-3)²(-1)
    -2(9)(-1)
    -18(-1)
    18

    8. x-(-2)

    9. subtract 7 then divide by -2

    10. 6+(-18)+(-13)+9
    -12-4
    -16

    11. 5x=11

    12. division property

    13. -3³ = -27

    14. x/2-3=7
    x/2=10
    x=20

    15. 2x-5=15

    16. x/-4-(-8)=12
    x/-4=4
    x=-16

    17. 168

    18. 16-20-(-8)-9
    -4-(-8)-9
    4-9
    -5 

  4. 1.B
    2.C
    3.D
    4.C
    5.C
    6.D

    That is the answers to the question that you asked and I hope that it helps you with the test.

  5. ANSWER TO QUESTION 1

    The given function is [tex]f(x)=-10x^2+4x[/tex]. We want to find the derivative of this function at [tex]x=11[/tex].

    The derivative of this function is given by,

    [tex]f'(x)=-20x+4[/tex]

    We now have to substitute  [tex]x=11[/tex] in to  [tex]f'(x)=-20x+4[/tex] to obtain,

    [tex]f'(11)=-20(11)+4[/tex]

    This implies that,

    [tex]f'(11)=-220+4[/tex]

    [tex]f'(11)=-216[/tex]

    Ans: A

    ANSWER TO Q2.

    The given limit is [tex]\lim_{x \to 4}^{ x^2+3x-1}[/tex].

    The function whose limit we are finding is a polynomial function. Since polynomial function are continuous everywhere, the limiting value is always equal to the functional value.

    Thus, [tex]\lim_{x \to 4}^{ x^2+3x-1}=f(4)[/tex]

    This implies that,

    Thus, [tex]\lim_{x \to 4}^{ x^2+3x-1}=(4)^2+3(4)-1[/tex].

    [tex]\lim_{x \to 4}^{ x^2+3x-1}=16+12-1[/tex]

    [tex]\lim_{x \to 4}^{ x^2+3x-1}=27[/tex]

    Ans: A.

    ANSWER TO Q3

    The given function is [tex]f(x)=7x+9[/tex].

    We want to find the derivative of this function at [tex]x=6[/tex].

    We must first of all differentiate this function to obtain,

    [tex]f'(x)=7[/tex]

    We can see that the derived function is constant, therefore value of [tex]x[/tex] will still give us [tex]7.[/tex]

    This implies that,

    [tex]f'(6)=7[/tex]

    Ans: A

    ANSWER TO Q4.

    See attachment

    ANSWER TO Q5.

    The given function is [tex]f(x)=\frac{3}{x}[/tex].

    To find the derivative of this function at,

    [tex]x=1[/tex], we must first differentiate this function.

    But let us rewrite the rational function as a power function so that it will be easier to differentiate using the power rule of differentiation.

    That is,

    [tex]f(x)=3x^{-1}[/tex].

    We differentiate now to obtain,

    [tex]f'(x)=-1(3x^{-1-1})[/tex]

    This implies that,

    [tex]f'(x)=-3x^{-2}[/tex]

    [tex]f'(x)=\frac{-3}{x^2}[/tex]

    At [tex]x=1[/tex]

    [tex]f'(1)=\frac{-3}{(1)^2}[/tex]

    [tex]f'(1)=-3[/tex]

    Ans: A

    ANSWER TO Q6

    The graph that has been described in the question is shown in the diagram above,

    We can see from the graph that as we approach [tex]x=2[/tex], from the left, the y-values are approaching [tex]4[/tex].

    That is,

    [tex]\lim_{x \to 2^{-}} f(x)=4[/tex]

    Also as we approach [tex]x=2[/tex] from the right, the y-values are approaching [tex]-3[/tex].

    That is,

    [tex]\lim_{x \to 2^{+}} f(x)=-3[/tex].

    Ans: D.

    ANSWER TO Q7

    The given piece-wise function is

    [tex]\lim_{x \to -1} \left \{ {{x+1,if\:x\:<\:-1} \atop {1-x,if\:x\:\ge -1}} \right.[/tex]

    Since [tex]x=-1 \in x\geq -1[/tex]

    We evaluate the limit at [tex]x=-1[/tex] of [tex]f(x)=1-x[/tex].

    Since this is a polynomial function, [tex]\lim_{x \to -1}f(x)= 1-x=f(-1)[/tex]

    This implies that,

    [tex]\lim_{x \to -1} f(x)=1-x=1--1[/tex]

    [tex]\lim_{x \to -1} f(x)=1-x=1--1[/tex]

    [tex]\lim_{x \to -1} f(x)=1-x=1+1[/tex]

    [tex]\lim_{x \to -1} f(x)=1-x=2[/tex]

    Ans: B.

    ANSWER TO Q8

    The given function is [tex]f(x)=\frac{1}{x-7}[/tex].

    This is a rational function that is defined for all real values except,[tex]x-7=0[/tex]

    Therefore the vertical asymptote is [tex]x=7[/tex]

    We can see from the graph above that, as x approaches seven from the left, the function approaches negative infinity

    That is [tex]\lim_{x \to 7^{-}} \frac{1}{x-7} =- \infty[/tex].

    Ans: A.

    ANSWER TO Q9

    The graph of the given piece-wise function is show as follows;

    We can see from the graph that, as the x-values are approaching 3 from the left y-values are approaching [tex]1.5[/tex].

    Note the limit is not the same as the functional value in this case. Also limit in this case is the y-value we are approaching as we get closer and closer to 3, not necessarily at 3.

    See graph

    Ans: B

    ANSWER TO Q10

    The given function is [tex]f(x)=\frac{x^3+1}{x^5}[/tex]

    We want to find the limiting value of this function as the x-values approaches zero.

    Thus,

    [tex]\lim_{x \to 0} f(x)=\frac{x^3+1}{x^5}[/tex]

    Since the function is not defined at [tex]x=0[/tex].

    We evaluate the one-sided limits as follows,

    [tex]\lim_{x \to 0^-} f(x)=\frac{x^3+1}{x^5}=- \infty[/tex]

    [tex]\lim_{x \to 0^+} f(x)=\frac{x^3+1}{x^5}=+ \infty[/tex]

    Since the right hand limit is not equal to the left hand limit,

    [tex]\lim_{x \to 0} f(x)=\frac{x^3+1}{x^5}[/tex] Does Not Exist.

    ANS: C

    ANSWER TO Q11

    The given function is

    [tex]f(x)=-\frac{3}{x}[/tex]

    To find the derivative of this function at,

    [tex]x=-4[/tex], we must first differentiate this function.

    But let us rewrite the rational function as a power function so that it will be easier to differentiate using the power rule of differentiation.

    That is,

    [tex]f(x)=-3x^{-1}[/tex].

    We differentiate now to obtain,

    [tex]f'(x)=-1(-3x^{-1-1})[/tex]

    This implies that,

    [tex]f'(x)=3x^{-2}[/tex]

    [tex]f'(x)=\frac{3}{x^2}[/tex]

    At [tex]x=-4[/tex]

    [tex]f'(1)=\frac{3}{(-4)^2}[/tex]

    [tex]f'(1)=\frac{3}{(16)}[/tex]

    Ans: B

    ANSWER TO Q12

    We want to find the limit of the function [tex]f(x)=x^2-2[/tex] as x approaches zero.

    [tex]\lim_{x \to 0} x^2-2[/tex]

    This is a polynomial function, therefore

    [tex]\lim_{x \to 0} x^2-2=f(0)=0^2-2=-2[/tex]

    Ans: B

    SEE  ATTACHMENT FOR ANSWER TO  QUESTIONS

    13, 14 and 15

     

    [tex]Question 1(multiple choice worth 6 points) find the derivative of f(x) = -10x2 + 4x at x = 11. -21[/tex]
    [tex]Question 1(multiple choice worth 6 points) find the derivative of f(x) = -10x2 + 4x at x = 11. -21[/tex]
    [tex]Question 1(multiple choice worth 6 points) find the derivative of f(x) = -10x2 + 4x at x = 11. -21[/tex]
    [tex]Question 1(multiple choice worth 6 points) find the derivative of f(x) = -10x2 + 4x at x = 11. -21[/tex]

  6. ANSWER TO QUESTION 1

    The given function is [tex]f(x)=-10x^2+4x[/tex]. We want to find the derivative of this function at [tex]x=11[/tex].

    The derivative of this function is given by,

    [tex]f'(x)=-20x+4[/tex]

    We now have to substitute  [tex]x=11[/tex] in to  [tex]f'(x)=-20x+4[/tex] to obtain,

    [tex]f'(11)=-20(11)+4[/tex]

    This implies that,

    [tex]f'(11)=-220+4[/tex]

    [tex]f'(11)=-216[/tex]

    Ans: A

    ANSWER TO Q2.

    The given limit is [tex]\lim_{x \to 4}^{ x^2+3x-1}[/tex].

    The function whose limit we are finding is a polynomial function. Since polynomial function are continuous everywhere, the limiting value is always equal to the functional value.

    Thus, [tex]\lim_{x \to 4}^{ x^2+3x-1}=f(4)[/tex]

    This implies that,

    Thus, [tex]\lim_{x \to 4}^{ x^2+3x-1}=(4)^2+3(4)-1[/tex].

    [tex]\lim_{x \to 4}^{ x^2+3x-1}=16+12-1[/tex]

    [tex]\lim_{x \to 4}^{ x^2+3x-1}=27[/tex]

    Ans: A.

    ANSWER TO Q3

    The given function is [tex]f(x)=7x+9[/tex].

    We want to find the derivative of this function at [tex]x=6[/tex].

    We must first of all differentiate this function to obtain,

    [tex]f'(x)=7[/tex]

    We can see that the derived function is constant, therefore value of [tex]x[/tex] will still give us [tex]7.[/tex]

    This implies that,

    [tex]f'(6)=7[/tex]

    Ans: A

    ANSWER TO Q4.

    See attachment

    ANSWER TO Q5.

    The given function is [tex]f(x)=\frac{3}{x}[/tex].

    To find the derivative of this function at,

    [tex]x=1[/tex], we must first differentiate this function.

    But let us rewrite the rational function as a power function so that it will be easier to differentiate using the power rule of differentiation.

    That is,

    [tex]f(x)=3x^{-1}[/tex].

    We differentiate now to obtain,

    [tex]f'(x)=-1(3x^{-1-1})[/tex]

    This implies that,

    [tex]f'(x)=-3x^{-2}[/tex]

    [tex]f'(x)=\frac{-3}{x^2}[/tex]

    At [tex]x=1[/tex]

    [tex]f'(1)=\frac{-3}{(1)^2}[/tex]

    [tex]f'(1)=-3[/tex]

    Ans: A

    ANSWER TO Q6

    The graph that has been described in the question is shown in the diagram above,

    We can see from the graph that as we approach [tex]x=2[/tex], from the left, the y-values are approaching [tex]4[/tex].

    That is,

    [tex]\lim_{x \to 2^{-}} f(x)=4[/tex]

    Also as we approach [tex]x=2[/tex] from the right, the y-values are approaching [tex]-3[/tex].

    That is,

    [tex]\lim_{x \to 2^{+}} f(x)=-3[/tex].

    Ans: D.

    ANSWER TO Q7

    The given piece-wise function is

    [tex]\lim_{x \to -1} \left \{ {{x+1,if\:x\:<\:-1} \atop {1-x,if\:x\:\ge -1}} \right.[/tex]

    Since [tex]x=-1 \in x\geq -1[/tex]

    We evaluate the limit at [tex]x=-1[/tex] of [tex]f(x)=1-x[/tex].

    Since this is a polynomial function, [tex]\lim_{x \to -1}f(x)= 1-x=f(-1)[/tex]

    This implies that,

    [tex]\lim_{x \to -1} f(x)=1-x=1--1[/tex]

    [tex]\lim_{x \to -1} f(x)=1-x=1--1[/tex]

    [tex]\lim_{x \to -1} f(x)=1-x=1+1[/tex]

    [tex]\lim_{x \to -1} f(x)=1-x=2[/tex]

    Ans: B.

    ANSWER TO Q8

    The given function is [tex]f(x)=\frac{1}{x-7}[/tex].

    This is a rational function that is defined for all real values except,[tex]x-7=0[/tex]

    Therefore the vertical asymptote is [tex]x=7[/tex]

    We can see from the graph above that, as x approaches seven from the left, the function approaches negative infinity

    That is [tex]\lim_{x \to 7^{-}} \frac{1}{x-7} =- \infty[/tex].

    Ans: A.

    ANSWER TO Q9

    The graph of the given piece-wise function is show as follows;

    We can see from the graph that, as the x-values are approaching 3 from the left y-values are approaching [tex]1.5[/tex].

    Note the limit is not the same as the functional value in this case. Also limit in this case is the y-value we are approaching as we get closer and closer to 3, not necessarily at 3.

    See graph

    Ans: B

    ANSWER TO Q10

    The given function is [tex]f(x)=\frac{x^3+1}{x^5}[/tex]

    We want to find the limiting value of this function as the x-values approaches zero.

    Thus,

    [tex]\lim_{x \to 0} f(x)=\frac{x^3+1}{x^5}[/tex]

    Since the function is not defined at [tex]x=0[/tex].

    We evaluate the one-sided limits as follows,

    [tex]\lim_{x \to 0^-} f(x)=\frac{x^3+1}{x^5}=- \infty[/tex]

    [tex]\lim_{x \to 0^+} f(x)=\frac{x^3+1}{x^5}=+ \infty[/tex]

    Since the right hand limit is not equal to the left hand limit,

    [tex]\lim_{x \to 0} f(x)=\frac{x^3+1}{x^5}[/tex] Does Not Exist.

    ANS: C

    ANSWER TO Q11

    The given function is

    [tex]f(x)=-\frac{3}{x}[/tex]

    To find the derivative of this function at,

    [tex]x=-4[/tex], we must first differentiate this function.

    But let us rewrite the rational function as a power function so that it will be easier to differentiate using the power rule of differentiation.

    That is,

    [tex]f(x)=-3x^{-1}[/tex].

    We differentiate now to obtain,

    [tex]f'(x)=-1(-3x^{-1-1})[/tex]

    This implies that,

    [tex]f'(x)=3x^{-2}[/tex]

    [tex]f'(x)=\frac{3}{x^2}[/tex]

    At [tex]x=-4[/tex]

    [tex]f'(1)=\frac{3}{(-4)^2}[/tex]

    [tex]f'(1)=\frac{3}{(16)}[/tex]

    Ans: B

    ANSWER TO Q12

    We want to find the limit of the function [tex]f(x)=x^2-2[/tex] as x approaches zero.

    [tex]\lim_{x \to 0} x^2-2[/tex]

    This is a polynomial function, therefore

    [tex]\lim_{x \to 0} x^2-2=f(0)=0^2-2=-2[/tex]

    Ans: B

    SEE  ATTACHMENT FOR ANSWER TO  QUESTIONS

    13, 14 and 15

     

    [tex]Question 1(multiple choice worth 6 points) find the derivative of f(x) = -10x2 + 4x at x = 11. -21[/tex]
    [tex]Question 1(multiple choice worth 6 points) find the derivative of f(x) = -10x2 + 4x at x = 11. -21[/tex]
    [tex]Question 1(multiple choice worth 6 points) find the derivative of f(x) = -10x2 + 4x at x = 11. -21[/tex]
    [tex]Question 1(multiple choice worth 6 points) find the derivative of f(x) = -10x2 + 4x at x = 11. -21[/tex]

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