All categories

3 years ago

Find (f/g)(x).Please help !!

Mathematics

1 answer:

Setler [38]3 years ago

6 0

Answer:D

Step-by-step explanation:

factorise 2x²-16x+32

x²-8x+16

(x-4)²

factorise 4x²-2x-20

2x²-x-10

2x²+4x-5x-10

2x(x+2)-5(x+2)

(2x-5)(x+2)

solving for (f/g)(x)

(x-4)²/(x+2)÷(2x-5)(x+2)/(x-4)²

(x-4)²/(x+2)*(x-4)²/(2x-5)(x+2)

(x-4)⁴/(2x-5)(x+2)²

You might be interested in

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago

Find the cost of 47.2 m cloth if the cost of 1 m cloth is $33.90

ICE Princess25 [194]

Answer: 1600.08

Step-by-step explanation:

47.2m*33.90

3 0

3 years ago

M - 37 = 56 jack has 56 marbles Kyle has 37 fewer than jack

Goshia [24]

Do we have to solve for M if so, the answer is

m-37 = 56

m =56 + 37

m = 93

hope it's helpful

8 0

3 years ago

Help me plz, i’d love u forever

Sergio039 [100]

1. We know that 1 feet = 12 inches

let x be a number of inches, which when added to 18 inches will give 7 feet

18 inches + x inches = 7 feet

18 inches + x inches = 84 inches [1 feet = 12 inches]

x inches = 84 inches - 18 inches

x inches = 66 inches

2. We know that 1 mile = 1760 yards

let x be the number of yards, when added to 2000 yards will be equal to 2 miles

2000 yards + x yards = 2 miles

2000 yards + x yards = 3520 yards

x yards = 1520 yards

3. We know that 1 feet = 12 inches

Number of feet = Given number of inches / 12

Number of feet = 108 / 12

Number of feet = 9 ft.

4. We know that 1 mile = 5280 feet

let x be the number of feet, when added to 8800 feet. will be equal to 8 miles

8800 feet + x feet = 8 miles

8800 feet + x feet = 42240 feet [1 mile = 5280 feet]

x feet = 42240 - 8800

x feet = 33440 ft.

5. We know that 12 inches = 1 feet

let x be the number of inches, when added to 37 inches, will be equal to 5 feet

37 inches + x inches = 5 feet

37 inches + x inches = 60 inches [12 inches = 1 feet]

x inches = 23 inches

6. We know that 1 yard = 3 feet

So, (number of yards)*3 = (number of feet)

Number of yards = number of feet / 3

Number of yards = 73 / 3 = 24.33 yards

7. We know that 1 yard = 3 feet

let x be the number of feet added to 15 feet to make 9 yards

15 feet + x feet = 9 yards

15 feet + x feet = 27 feet [1 yard = 3 feet]

x feet = 12 ft

8. We know that 1 mile = 1760 yards

let x be the number of yards remaining to reach 4 miles

3000 yards + x yards = 4 miles

3000 yards + x yards = 7040

x yards = 4040 yards

9. We know that 1 yard = 3 feet

So, number of yards * 3 = number of feet

replacing the values:

number of yards*3 = 114

number of yards = 114/3

number of yards = 38 yards

10. We know that 1 mile = 5280 feet

let x be the number of feet that must be added to 9000 feet to make 3 miles

x feet + 9000 feet = 3 miles

x feet + 9000 feet = 15840 feet

x ft = 6840 ft

11. We know that 12 inches = 1 feet

let x be the number of inches that must be added to 46 inches to get a length of 9 feet

46 inches + x inches = 9 feet

46 inches + x inches = 108 inches

x inches = 108 - 46 inches

x inches = 62 inches

12. We know that 1 mile = 1760 yards

let x be the number of yards that must be added to 3500 yards to make 2 miles

3500 yards + x yards = 2 miles

3500 yards + x yards = 3520 yards

x = 20 yards

13. We know that 1 ft = 12 inches

So, number of feet * 12 = number of inches

number of feet = number of inches / 12

number of feet in 252 inches

Number of feet = 252/12

Number of feet = 21 ft

14. We know that 1 mile = 5280 feet

let x be the number of feet, when added to 6500 feet, will be equal to 2 miles

x ft + 6500 ft = 2 miles

x ft + 6500 ft = 10560 ft

x ft = 4060 ft

15. We know that 1 mile = 5280 feet

let x be the number of feet, when added to 4500 feet, will be equal to 1 mile

x ft + 4500 ft = 1 mile

x ft + 4500 ft = 5280 ft

x ft = 780 ft

3 0

3 years ago

Mary must choose a new password where the first and last choices are possibly repeated lowercase letters; the second and third

Salsk061 [2.6K]

The number of ways she can choose a password is 3986236800

<h3>In how many ways can she choose a password?</h3>

The given parameters and the possible selection of characters are:

First and last choices are possibly repeated lowercase letters;

There are 26 lower characters.

Since the characters can be repeated, then we have

First = 26

Last = 26

The second and third positions must be distinct uppercase letters

There are 26 upper characters.

Since the characters are distinct, then we have

Second = 26

Third = 25

The fourth position must be a # , $, or & symbol;

So, we have

Fourth = 3

The next four positions are distinct nonzero digits.

There are 9 nonzero digits.

Since the digits are distinct, then we have

Next = 9, 8, 7, 6

The number of ways she can choose a password is

Ways = First * Second * Third * Fourth * Next * Last

So, we have

Ways = 26 * 26 * 25 * 3 * 9 * 8 * 7 * 6 * 26

Evaluate the product

Ways = 3986236800

Hence, the number of ways she can choose a password is 3986236800