4^5 x 4^-7 divided by 4^-2, №18010243, 18.02.2023 18:04

Mathematics

4^5 x 4^-7 divided by 4^-2

Step-by-step answer

P Answered by Specialist

Step-by-step explanation:

Our expression;

$\frac{4^5\times4^7}{4^-2}$

Notes;

When exponents multiply, they add.

When exponents divide, they subtract.

Multiply;

$\frac{4^5\times4^7}{4^{-2}}=\frac{4^{-2}}{4^{-2}}$

Divide;

$\boxed{\frac{4^{-2}}{4^{-2}}=4^0=1}$

Mathematics

Which expression is equal to 4^5 x 4^-7 divided by 4^-2? help asap

Step-by-step answer

P Answered by Master

7

4^0

Step-by-step explanation:

whenever the power is to 0 the answer is 1

4^5 x 4^-7 ÷ 4^-2 = 1

4^0 = 1

Mathematics

Which expression is equal to 4^5 x 4^-7 divided by 4^-2

Step-by-step answer

P Answered by Master

17

1

Step-by-step explanation:

4^5 * 4^-7 = 4^-2

4^-2 / 4^-2 = 4^0

4^0 = 1

Mathematics

Which expression is equal to 4^5 x 4^-7 divided by 4^-2? help asap

Step-by-step answer

P Answered by Specialist

7

4^0

Step-by-step explanation:

whenever the power is to 0 the answer is 1

4^5 x 4^-7 ÷ 4^-2 = 1

4^0 = 1

Mathematics

7.06 1. find the first six terms of the sequence. a1 = -7, an = 4 • an-1 a) -7, -28, -112, -448, -1792, -7168 b) -28, -112, -448, -1792, -7168, -28,672 c) -7, -28, -24, -20, -16, -12 d) 0, 4, -28, -24, -20, -16 2. find an equation for the nth term of the arithmetic sequence. -13, -8, -3, 2, an = -13 x 5(n - 1) an = -13 + 5(n - 1) an = -13 + 5(n + 2) an = -13 + 5(n + 1) 3. find an equation for the nth term of the arithmetic sequence. a15 = -53, a16 = -5 a) an = -725 - 48(n - 1) b) an = -725 + 48(n + 1) c) an = -725 + 48(n - 1) d) an = -725 - 48(n

Step-by-step answer

P Answered by PhD

21

1. A. According to the expression a_n=4*a_n-1, each term after a1 is four times the previous term. The first term is -7 as given, 2nd term should be -7*4=-28, 3rd term is -28*4=-112, ... A is the correct answer.

2. B. The sequence is -13, -8, -3, 2... It's obvious that each term is equal to the previous term plus 5. This is an arithmetic sequence with initial term -13 and common difference 5. We know a1=-13, so a_n=-13+5*(n-1). The answer is B.

3. A. We are given a15=-53, a16=-5. The common difference of the arithmetic sequence is -5-(-53)=48. The formula for a_n term is a1+48*(n-1). We know that a15=-13; plug in n=15, a15=-53=a1+48*(15-1), a1=-725. So a_n=-725+48*(n-1).

4. Diverge. We are given a few terms, 11, 44, 176, 704... Observe that each term is four times the previous one. 11*4=44, 44*4=176, 176*4=704... This is a geometric series with common ratio>1. You can keep multiplying by 4 and the series goes to infinity, so it diverges.

5. D. We have -4, -16, -64, -256... Same as above, each term is four times the previous one. The initial term is a1=-4. The common ratio d=4. So a_n=a1*d^(n-1)=-4*4^(n-1)=-4^n. (D).

6. The answer is A. a2=-2, a5=16. Suppose the common ratio is D. a_n=a1*d^(n-1). a2=a1*d; a5=a1*d^4. Plug in a2 and a5: -2=a1*d, 16=a1*d^4. 16/-2=d^3=-8, d=-2, a1=1. So a_n=1*(-2)^(n-1).

7. B. We are given the sequence 4, -24, 144,... Each term is -6 times the previous one. The first term a0=4, the n^th term a_(n-1) is a1*d^n=4*(-6)^n. To express the sum, we simply have to use the sigma notation and sum 4*(-6)^n from n=0 to infinity. The answer is B.

8. D. We are given -3 + 6 + 15 + 24... 132. Each term is equal to the previous one plus 9. First term a0=-3, n^th term a_n-1 is -3+9*n. The last term is 132. 132 =-3+9n, n=15. So we have to sum -3+9n from n=0 to n=15.

9. B. 343 + 512 + 729 + 1000+... 343=7^3, 512=8^3, 729=9^3, 1000=10^3. This is a sequence of perfect cubes. Therefore, the sum is n^3 from n=7 to infinity. (The initial term is 343=7^3).

10. B. We are given 3, 5, 7, 9, ... 21. The common difference is 2. There are (21-3)/2+1=10 terms. The initial term a1=3, and last term is a10=21. The sum is (a1+a10)*10/2=(3+21)*10/2=120.

11. C. 4/3, 16/3, 64/3, 256/3, 1024/3. Each term is four times the previous one. This is a geometric series with initial term a1=4/3 and common ratio r=4. 1024/3 is the 5th term of the sequence. So sum=a1*(1-r^n)/(1-r)=4/3*(1-4^5)/(1-4)=-4/9*-1023=1364/3.

12. B. 10,12,14,... This is an arithmetic sequence. a1=10, and common difference d=2. There are 20 terms (20 rows). a20=a1+d*(n-1)=10+2*(20-1)=48. So the sum S=(a1+an)*n/2=(10+48)*20/2=580.

13. 10 + 20 + 30 + ... + 10n = 5n(n + 1). When n=1, this expression is true, since 10=5*1*(1+1). Suppose when n=k, this statement is true, then when n=k+1, the left side is 10+...+10n+10(n+1), the right side is 5(n+1)(n+2). The left side adds 10(n+1) compared to the previous one. The right side adds 5(n+1)(n+2)-5n(n+1)=5(n+1)(n+2-n)=10(n+1). So the statement holds true.

14. The height at week 0 is a0=300 (initial height). Common difference is 4.2 (weekly increment). a_n=300+4.2n. At week n, the height of the tree is 300+4.2*n centimeters.

Mathematics

7.06 1. find the first six terms of the sequence. a1 = -7, an = 4 • an-1 a) -7, -28, -112, -448, -1792, -7168 b) -28, -112, -448, -1792, -7168, -28,672 c) -7, -28, -24, -20, -16, -12 d) 0, 4, -28, -24, -20, -16 2. find an equation for the nth term of the arithmetic sequence. -13, -8, -3, 2, an = -13 x 5(n - 1) an = -13 + 5(n - 1) an = -13 + 5(n + 2) an = -13 + 5(n + 1) 3. find an equation for the nth term of the arithmetic sequence. a15 = -53, a16 = -5 a) an = -725 - 48(n - 1) b) an = -725 + 48(n + 1) c) an = -725 + 48(n - 1) d) an = -725 - 48(n

Step-by-step answer

P Answered by PhD

21

1. A. According to the expression a_n=4*a_n-1, each term after a1 is four times the previous term. The first term is -7 as given, 2nd term should be -7*4=-28, 3rd term is -28*4=-112, ... A is the correct answer.

2. B. The sequence is -13, -8, -3, 2... It's obvious that each term is equal to the previous term plus 5. This is an arithmetic sequence with initial term -13 and common difference 5. We know a1=-13, so a_n=-13+5*(n-1). The answer is B.

3. A. We are given a15=-53, a16=-5. The common difference of the arithmetic sequence is -5-(-53)=48. The formula for a_n term is a1+48*(n-1). We know that a15=-13; plug in n=15, a15=-53=a1+48*(15-1), a1=-725. So a_n=-725+48*(n-1).

4. Diverge. We are given a few terms, 11, 44, 176, 704... Observe that each term is four times the previous one. 11*4=44, 44*4=176, 176*4=704... This is a geometric series with common ratio>1. You can keep multiplying by 4 and the series goes to infinity, so it diverges.

5. D. We have -4, -16, -64, -256... Same as above, each term is four times the previous one. The initial term is a1=-4. The common ratio d=4. So a_n=a1*d^(n-1)=-4*4^(n-1)=-4^n. (D).

6. The answer is A. a2=-2, a5=16. Suppose the common ratio is D. a_n=a1*d^(n-1). a2=a1*d; a5=a1*d^4. Plug in a2 and a5: -2=a1*d, 16=a1*d^4. 16/-2=d^3=-8, d=-2, a1=1. So a_n=1*(-2)^(n-1).

7. B. We are given the sequence 4, -24, 144,... Each term is -6 times the previous one. The first term a0=4, the n^th term a_(n-1) is a1*d^n=4*(-6)^n. To express the sum, we simply have to use the sigma notation and sum 4*(-6)^n from n=0 to infinity. The answer is B.

8. D. We are given -3 + 6 + 15 + 24... 132. Each term is equal to the previous one plus 9. First term a0=-3, n^th term a_n-1 is -3+9*n. The last term is 132. 132 =-3+9n, n=15. So we have to sum -3+9n from n=0 to n=15.

9. B. 343 + 512 + 729 + 1000+... 343=7^3, 512=8^3, 729=9^3, 1000=10^3. This is a sequence of perfect cubes. Therefore, the sum is n^3 from n=7 to infinity. (The initial term is 343=7^3).

10. B. We are given 3, 5, 7, 9, ... 21. The common difference is 2. There are (21-3)/2+1=10 terms. The initial term a1=3, and last term is a10=21. The sum is (a1+a10)*10/2=(3+21)*10/2=120.

11. C. 4/3, 16/3, 64/3, 256/3, 1024/3. Each term is four times the previous one. This is a geometric series with initial term a1=4/3 and common ratio r=4. 1024/3 is the 5th term of the sequence. So sum=a1*(1-r^n)/(1-r)=4/3*(1-4^5)/(1-4)=-4/9*-1023=1364/3.

12. B. 10,12,14,... This is an arithmetic sequence. a1=10, and common difference d=2. There are 20 terms (20 rows). a20=a1+d*(n-1)=10+2*(20-1)=48. So the sum S=(a1+an)*n/2=(10+48)*20/2=580.

13. 10 + 20 + 30 + ... + 10n = 5n(n + 1). When n=1, this expression is true, since 10=5*1*(1+1). Suppose when n=k, this statement is true, then when n=k+1, the left side is 10+...+10n+10(n+1), the right side is 5(n+1)(n+2). The left side adds 10(n+1) compared to the previous one. The right side adds 5(n+1)(n+2)-5n(n+1)=5(n+1)(n+2-n)=10(n+1). So the statement holds true.

14. The height at week 0 is a0=300 (initial height). Common difference is 4.2 (weekly increment). a_n=300+4.2n. At week n, the height of the tree is 300+4.2*n centimeters.

Mathematics

Need answers as soon as possible, will give brainiest and rate 5/5. 1: solve x/-3 = -15. 45 -45 5 -5 2: find the value of -7 + 3(-12) ÷ (-3). -16 -19 -40 5 3: if f(n) = n^2 - n, then f(-4) is -20 20 12 -12 4: which equation does not have the same solution as the others? x/4 = 2 x - 9 = 17 x + 12 = 20 2x = 16 5: the solution to 2x - 5 = 27 is also a solution of which of the following equations? 3 + 5x = 58 3x - 2 = 31 2x + 3 = 35 27 - 2x = 5 6: find the value of -8 - 12 - (-20). 24 -40 16 0 7: evaluate -2x ^2y, if x = -3 and y = -1. -12 -18 18 12

Step-by-step answer

P Answered by Master

54

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

Mathematics

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is an expression for the length of the pool? a. 2x^2+8x+3 b. 2x^2-8x-3 c. 2x^2-8x+3 d. 2x^2+8-3 2. simplify x/6x-x^2 a. 1/6-x; where x = 0,6 b. 1/6-x; where x=6 c. 1/6; where x=0 d. 1/6 3. simplify -12x4/x4+8x^5 a. -12/1+8x; where x= -1/8 b. -12/1+8x; where x= -1/8,0 4. simplify x+5/ x^2+6x+5 a. 1/x+1; where x= -1 b. 1/x+1; where x=-1, -5 5. simplify x^2-3x-18/x+3 a. x-3 b. x-6; where x= -3 c. x-6; where x= 6 d. 1/x+3; where x= -3 6. simplify 2/3a .

Step-by-step answer

P Answered by PhD

20

Question 1

To find the width of the rectangle, we divide the area by the length
$2x^{3}-29x+12$ ÷ x+4

We use the method of long division to get the answer. The method is shown in the first diagram below

$2x^{2}-8x+3$

Question 2:
$\frac{x}{6x-x^{2} } = \frac{x}{x(6-x)} = \frac{1}{6-x}$

Question 3:
$\frac{-12 x^{4} }{x^{4}+8 x^{5} }= \frac{-12 x^{4} }{ x^{4}(1+8x)}= \frac{-12}{1+8x}$

Question 4:
$\frac{x+5}{x^{2}+6x+5}= \frac{x+5}{(x+1)(x+5)}= \frac{1}{x+1}$

Question 5:
$\frac{x^{2}-3x-18} {x+3}= \frac{(x-6)(x+3)}{x+3}= \frac{x-6}{1}=x-6$

Question 6:
$\frac{2}{3a}$ × $\frac{2}{a^{2}}$ = $\frac{4}{3a^{3} }$ where $a \neq 0$

Question 7: (Question is not written well)
$\frac{x-5}{4x+8}$ × $(12x^{2}+32x+8)$
$\frac{12 x^{3}-28 x^{2} -152x-40 }{4x+8}$
By performing long division we get an answer $3 x^{2} -x-36$ with remainder of 248

Question 8:
$( \frac{x^{2}-16} {x-1})$ ÷ (x+4)

$( \frac{ x^{2}-16 }{x-1})$ × $\frac{1}{x+4}$
$\frac{(x+4)(x-1)}{x-1}$ × $\frac{1}{x+4}$
Cancelling out x+4

we obtain $\frac{x+1}{x-1}$

Question 9:
$\frac{x^{2}+2x+1} {x-2}$ ÷ $\frac{x^{2-1} }{x^{2}-4 }$
$\frac{ x^{2}+2x+1 }{x-2}$ × $\frac{x^{2}-4 }{x^{2}-1}$
Factorise all the quadratic expression gives
$\frac{(x+1)(x+1)}{x-2}$ × $\frac{(x-2)(x+2)}{(x+1)(x-1)}$
Cancelling out (x+1)

and

gives a simplest form
$\frac{(x+1)(x+2)}{x-1}$

Question 10:

$\frac{24 w^{10}+8w^{12} }{4 x^{4} }= \frac{24w^{10} }{4 x^{4} } + \frac{8 w^{12} }{4 x^{4} }$
Cancelling out the constants of each fraction
$\frac{6w^{10} }{x^{4} }+ \frac{2w^{12} }{x^{4}}= \frac{6w^{10}+2w^{12} }{ x^{4}}$

Question 11:

$\frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} } = \frac{-2m^{6}(3m^{3}-3m^{2}-8)}{2m^{3} }$
Cancelling $2m^{3}$ gives us the simplified form
$-m^{3}(3m^{3}-3m^{2}-8) = -3m^{6}+3m^{5}+8m^{3}$

Question 12:

$\frac{-4x}{x+7} - \frac{8}{x-7} = \frac{-4x(x-7)-8(x+7)}{(x+7)(x-7)}$
$\frac{-4 x^{2} +28x-8x-56}{(x+7)(X-7)}= \frac{-4 x^{2} +20x-56}{(x+7)(x-7)}$
Factorising the numerator expression
$\frac{(-4x+28)(x-2)}{(x+7)(x-7)} = \frac{-4(x-7)(x-2)}{(x+7)(x-7)}$
Cancelling out x-7

gives the simplified form
$\frac{-4x+8}{x-7}$

Question 13:

$\frac{3}{x-3} - \frac{5}{x-2}= \frac{x3(x-2)-5(x-2)}{y(x-3)(x-2)}$
$\frac{3x-6-5x+15}{(x-3)(x-2)}= \frac{-2x+9}{(x-3)(x-2)}$

Question 14:

$\frac{9}{x-1}- \frac{5}{x+4}= \frac{9(x+4)-5(x-1)}{(x-1)(x+4)}$ $\frac{9x+36-5x+5}{(x-1)(x+4)}= \frac{4x+41}{(x-1)(x+4)}$

Question 15:

$\frac{-3}{x+2}- \frac{(-5)}{x+3}= \frac{-3(x+3)-(-5)(x+2)}{(x+2)(x+3)}$
$\frac{-3x-9+5x+10}{(x+2)(x+3)}= \frac{2x+1}{(x+2)(x+3)}$

Question 16:

$\frac{4}{x}+ \frac{5}{x}=-3$
$\frac{9}{x}=-3$
x=-3

Question 17:

$\frac{1}{3x-6}- \frac{5}{x-2}=12$
$\frac{(x-2)-5(3x-6)}{(3x-6)(x-2)} = \frac{x-2-15x+30}{(3x-6)(x-2)}= \frac{-14x+28}{(3x-6)(x-2)}$

Question 18

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is

Mathematics

Need answers as soon as possible, will give brainiest and rate 5/5. 1: solve x/-3 = -15. 45 -45 5 -5 2: find the value of -7 + 3(-12) ÷ (-3). -16 -19 -40 5 3: if f(n) = n^2 - n, then f(-4) is -20 20 12 -12 4: which equation does not have the same solution as the others? x/4 = 2 x - 9 = 17 x + 12 = 20 2x = 16 5: the solution to 2x - 5 = 27 is also a solution of which of the following equations? 3 + 5x = 58 3x - 2 = 31 2x + 3 = 35 27 - 2x = 5 6: find the value of -8 - 12 - (-20). 24 -40 16 0 7: evaluate -2x ^2y, if x = -3 and y = -1. -12 -18 18 12

Step-by-step answer

P Answered by Master

54

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

Mathematics

Question 1(multiple choice worth 6 points) find the derivative of f(x) = -10x2 + 4x at x = 11. -216 -196 -176 -363 question 2(multiple choice worth 6 points) find the limit of the function by using direct substitution. limit as x approaches four of quantity x squared plus three x minus one does not exist -27 0 27 question 3(multiple choice worth 6 points) find the derivative of f(x) = 7x + 9 at x = 6. 7 9 6 0 question 4(multiple choice worth 7 points) find the indicated limit, if it exists. limit of f of x as x approaches 0 where f of x equals

Step-by-step answer

P Answered by PhD

2

ANSWER TO QUESTION 1

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

ANSWER TO Q2.

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.

ANSWER TO Q3

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 will still give us

This implies that,

f'(6)=7

Ans: A

ANSWER TO Q4.

See attachment

ANSWER TO Q5.

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

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

That is,

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

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

That is,

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

Ans: D.

ANSWER TO Q7

The given piece-wise function is

$\lim_{x \to -1} \left \{ {{x+1,if\:x\:$

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,

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

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

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

Ans: B.

ANSWER TO Q8

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.

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 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

ANSWER TO Q10

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