# 5/6 – 1/3 divided by 5/12

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

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

## This Post Has 10 Comments

1. joejoefofana says:

1=B
2=C
3=D
4=A
5=C
6=D
JUTKD

2. AquariusOx says:

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

3. jdkrisdaimcc11 says:

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

5. koolja3 says:

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

7. skent1423 says:

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.

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

The derivative of this function is given by,

$f'(x)=-20x+4$

We now have to substitute  $x=11$ in to  $f'(x)=-20x+4$ to obtain,

$f'(11)=-20(11)+4$

This implies that,

$f'(11)=-220+4$

$f'(11)=-216$

Ans: A

The given limit is $\lim_{x \to 4}^{ x^2+3x-1}$.

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, $\lim_{x \to 4}^{ x^2+3x-1}=f(4)$

This implies that,

Thus, $\lim_{x \to 4}^{ x^2+3x-1}=(4)^2+3(4)-1$.

$\lim_{x \to 4}^{ x^2+3x-1}=16+12-1$

$\lim_{x \to 4}^{ x^2+3x-1}=27$

Ans: A.

The given function is $f(x)=7x+9$.

We want to find the derivative of this function at $x=6$.

We must first of all differentiate this function to obtain,

$f'(x)=7$

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

This implies that,

$f'(6)=7$

Ans: A

See attachment

The given function is $f(x)=\frac{3}{x}$.

To find the derivative of this function at,

$x=1$, 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,

$f(x)=3x^{-1}$.

We differentiate now to obtain,

$f'(x)=-1(3x^{-1-1})$

This implies that,

$f'(x)=-3x^{-2}$

$f'(x)=\frac{-3}{x^2}$

At $x=1$

$f'(1)=\frac{-3}{(1)^2}$

$f'(1)=-3$

Ans: A

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 $x=2$, from the left, the y-values are approaching $4$.

That is,

$\lim_{x \to 2^{-}} f(x)=4$

Also as we approach $x=2$ from the right, the y-values are approaching $-3$.

That is,

$\lim_{x \to 2^{+}} f(x)=-3$.

Ans: D.

The given piece-wise function is

$\lim_{x \to -1} \left \{ {{x+1,if\:x\:<\:-1} \atop {1-x,if\:x\:\ge -1}} \right.$

Since $x=-1 \in x\geq -1$

We evaluate the limit at $x=-1$ of $f(x)=1-x$.

Since this is a polynomial function, $\lim_{x \to -1}f(x)= 1-x=f(-1)$

This implies that,

$\lim_{x \to -1} f(x)=1-x=1--1$

$\lim_{x \to -1} f(x)=1-x=1--1$

$\lim_{x \to -1} f(x)=1-x=1+1$

$\lim_{x \to -1} f(x)=1-x=2$

Ans: B.

The given function is $f(x)=\frac{1}{x-7}$.

This is a rational function that is defined for all real values except,$x-7=0$

Therefore the vertical asymptote is $x=7$

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

That is $\lim_{x \to 7^{-}} \frac{1}{x-7} =- \infty$.

Ans: A.

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 $1.5$.

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

The given function is $f(x)=\frac{x^3+1}{x^5}$

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

Thus,

$\lim_{x \to 0} f(x)=\frac{x^3+1}{x^5}$

Since the function is not defined at $x=0$.

We evaluate the one-sided limits as follows,

$\lim_{x \to 0^-} f(x)=\frac{x^3+1}{x^5}=- \infty$

$\lim_{x \to 0^+} f(x)=\frac{x^3+1}{x^5}=+ \infty$

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

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

ANS: C

The given function is

$f(x)=-\frac{3}{x}$

To find the derivative of this function at,

$x=-4$, 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,

$f(x)=-3x^{-1}$.

We differentiate now to obtain,

$f'(x)=-1(-3x^{-1-1})$

This implies that,

$f'(x)=3x^{-2}$

$f'(x)=\frac{3}{x^2}$

At $x=-4$

$f'(1)=\frac{3}{(-4)^2}$

$f'(1)=\frac{3}{(16)}$

Ans: B

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

$\lim_{x \to 0} x^2-2$

This is a polynomial function, therefore

$\lim_{x \to 0} x^2-2=f(0)=0^2-2=-2$

Ans: B

SEE  ATTACHMENT FOR ANSWER TO  QUESTIONS

13, 14 and 15

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

9. belle200163 says:

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

The derivative of this function is given by,

$f'(x)=-20x+4$

We now have to substitute  $x=11$ in to  $f'(x)=-20x+4$ to obtain,

$f'(11)=-20(11)+4$

This implies that,

$f'(11)=-220+4$

$f'(11)=-216$

Ans: A

The given limit is $\lim_{x \to 4}^{ x^2+3x-1}$.

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, $\lim_{x \to 4}^{ x^2+3x-1}=f(4)$

This implies that,

Thus, $\lim_{x \to 4}^{ x^2+3x-1}=(4)^2+3(4)-1$.

$\lim_{x \to 4}^{ x^2+3x-1}=16+12-1$

$\lim_{x \to 4}^{ x^2+3x-1}=27$

Ans: A.

The given function is $f(x)=7x+9$.

We want to find the derivative of this function at $x=6$.

We must first of all differentiate this function to obtain,

$f'(x)=7$

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

This implies that,

$f'(6)=7$

Ans: A

See attachment

The given function is $f(x)=\frac{3}{x}$.

To find the derivative of this function at,

$x=1$, 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,

$f(x)=3x^{-1}$.

We differentiate now to obtain,

$f'(x)=-1(3x^{-1-1})$

This implies that,

$f'(x)=-3x^{-2}$

$f'(x)=\frac{-3}{x^2}$

At $x=1$

$f'(1)=\frac{-3}{(1)^2}$

$f'(1)=-3$

Ans: A

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 $x=2$, from the left, the y-values are approaching $4$.

That is,

$\lim_{x \to 2^{-}} f(x)=4$

Also as we approach $x=2$ from the right, the y-values are approaching $-3$.

That is,

$\lim_{x \to 2^{+}} f(x)=-3$.

Ans: D.

The given piece-wise function is

$\lim_{x \to -1} \left \{ {{x+1,if\:x\:<\:-1} \atop {1-x,if\:x\:\ge -1}} \right.$

Since $x=-1 \in x\geq -1$

We evaluate the limit at $x=-1$ of $f(x)=1-x$.

Since this is a polynomial function, $\lim_{x \to -1}f(x)= 1-x=f(-1)$

This implies that,

$\lim_{x \to -1} f(x)=1-x=1--1$

$\lim_{x \to -1} f(x)=1-x=1--1$

$\lim_{x \to -1} f(x)=1-x=1+1$

$\lim_{x \to -1} f(x)=1-x=2$

Ans: B.

The given function is $f(x)=\frac{1}{x-7}$.

This is a rational function that is defined for all real values except,$x-7=0$

Therefore the vertical asymptote is $x=7$

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

That is $\lim_{x \to 7^{-}} \frac{1}{x-7} =- \infty$.

Ans: A.

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 $1.5$.

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

The given function is $f(x)=\frac{x^3+1}{x^5}$

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

Thus,

$\lim_{x \to 0} f(x)=\frac{x^3+1}{x^5}$

Since the function is not defined at $x=0$.

We evaluate the one-sided limits as follows,

$\lim_{x \to 0^-} f(x)=\frac{x^3+1}{x^5}=- \infty$

$\lim_{x \to 0^+} f(x)=\frac{x^3+1}{x^5}=+ \infty$

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

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

ANS: C

The given function is

$f(x)=-\frac{3}{x}$

To find the derivative of this function at,

$x=-4$, 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,

$f(x)=-3x^{-1}$.

We differentiate now to obtain,

$f'(x)=-1(-3x^{-1-1})$

This implies that,

$f'(x)=3x^{-2}$

$f'(x)=\frac{3}{x^2}$

At $x=-4$

$f'(1)=\frac{3}{(-4)^2}$

$f'(1)=\frac{3}{(16)}$

Ans: B

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

$\lim_{x \to 0} x^2-2$

This is a polynomial function, therefore

$\lim_{x \to 0} x^2-2=f(0)=0^2-2=-2$

Ans: B

SEE  ATTACHMENT FOR ANSWER TO  QUESTIONS

13, 14 and 15

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

10. ghimiren7443 says:

6.6. ans. is 2
21. and is