All categories

3 years ago

Plz solve this question

Mathematics

1 answer:

bixtya [17]3 years ago

4 0

Answer:

$\frac{2(x - 4)}{(x - 2)(x + 2)}$

Step-by-step explanation:

$\frac{2}{(x - 2)} - \frac{8}{ ({x}^{2} - 4 )} \\ \frac{2}{(x - 2)} - \frac{8}{(x - 2)(x + 2)} \\ \frac{2(x + 2) - 8}{(x - 2)(x + 2)} \\ \frac{2x + 4 - 8}{(x - 2)(x + 2)} \\ \frac{2x - 8}{(x - 2)(x + 2)} \\ \frac{2(x - 4)}{(x - 2)(x + 2)}$

You might be interested in

What's the radius of a cylinder with a volume of 150 in cubed and a height of 10 in

SOVA2 [1]

Answer:

2.19

Step-by-step explanation:

7 0

2 years ago

Express the following

krok68 [10]

Answer:

1. 32

2. 24

3. $\frac{1}{12}$

4. $\frac{1}{8}$

5. 15

6. $\frac{1}{15}$

Step-by-step explanation:

1. 4÷ $\frac{1}{8} = 4*8=32$

2. 6÷ $\frac{1}{4} = 6*4=24$

3. $\frac{1}{3}$ ÷4= $\frac{1}{3}*\frac{1}{4} =\frac{1}{12}$

4. $\frac{1}{2}$ ÷4= $\frac{1}{2} *\frac{1}{4} =\frac{1}{8}$

5. 5÷ $\frac{1}{3} =5*3=15$

6. $\frac{1}3}$ ÷5= $\frac{1}{3} *\frac{1}{5} = \frac{1}{15}$

3 0

3 years ago

if a, b, and c are prime numbers, do (a b) and c have a common factor that is greater than 1? (1) a, b, and c are all different

klio [65]

If a,b, and c are prime numbers, do (a*b) and c have a common factor that is greater than 1

(1) a,b, and c are all different prime numbers

(2) c≠2

1. Let's assume values of 1,3 and 5 to a, b, and c respectively

a b = 1*3 = 3

3 and 5 do not have any common factor aside 1

Let's assume values of 1,3 and 2 to a,b and c respectively

a *b = 1*3 = 3

3 and 2 does not have a common factor aside 1

2. c $\neq$ 2

Let's assume values of 2,7 and 3 to a b and c response

a * b = 2 *7 = 14

14 and 3 does not have a common factor aside 1

learn more about of prime number here

brainly.com/question/14410795

#SPJ4

8 0

11 months ago

Determine the inverse of the function. (show work please)

uysha [10]

Answer:

This is easy -- it's just a list of steps. At this level, the problems are pretty simple.

Let's just do one, then I'll write out the list of steps for you.

Find the inverse of f( x ) = -( 1 / 3 )x + 1

STEP 1: Stick a "y" in for the "f(x)" guy:

y = -( 1 / 3 )x + 1

STEP 2: Switch the x and y

( because every (x, y) has a (y, x) partner! ):

x = -( 1 / 3 )y + 1

STEP 3: Solve for y:

x = -( 1 / 3 )y + 1 ... multiply by 3 to ditch the fraction ... 3x = -y + 3 ... ditch the +3 ... subtract 3 from both sides ... 3x - 3 = -y ... multiply by -1 ... -3x + 3 = y ... y = -3x + 3

STEP 4: Stick in the inverse notation, f^( -1 )( x )

f^( -1 )( x ) = -3x + 3

Step-by-step explanation:

5 0

3 years ago

Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,

posledela

Answer:

number of multiplies is n!
n=10, 3.6 ms
n=15, 21.8 min
n=20, 77.09 yr
n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

mpy[n] = n·mp[n-1]

mpy[2] = 2

So, ...

mpy[n] = n! . . . n ≥ 2

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0

3 years ago

Plz solve this question​

Plz solve this question