5/6 - 1/3 divided by 5/12

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5/6 – 1/3 divided by 5/12

5/6 - 1/3 divided by 5/12

[tex]5/6 - 1/3 divided by 5/12[/tex]

1=B

2=C

3=D

4=A

5=C

6=D

JUTKD

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

5 songs and I have a few more songs and more to be able

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

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

1. x+9

2. (b-5)+6

3. 15+5v

4. 2/3k

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.

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]

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.

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]

6.6. ans. is 2

21. and is