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xxTIMURxx [149]
3 years ago
12

Please help I’ll give 50 points

Mathematics
2 answers:
tiny-mole [99]3 years ago
6 0

When dividing roots, multiply them together: 4 x 5 = 20

Invert the product: 1/20

Raise the number under the root by that

Answer = 6 ^ 1/20

DIA [1.3K]3 years ago
4 0

Answer:

C

Step-by-step explanation:

(4√6)/(5√6)

= 6^(¼) ÷ 6^(1/5)

= 6^(1/4 - 1/5)

= 6^[(5-4)/20]

= 6^(1/20

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Express 2x+1/(x-2)(x²+1) as a partial fraction.
Oduvanchick [21]

Answer:

Partial fraction = 1/(x-2) - x/(x^2+1)

Step-by-step explanation:

Question:

Express 2x+1/(x-2)(x²+1) as a partial fraction.

Note: it will be assumed that there was a typo in the interpretation of parentheses to mean

(2x+1) / ( (x-2)(x^2+1) )

Let

(2x+1) / ( (x-2)(x^2+1) ) = A/(x-2) + (Bx+C)/(x^2+1) .........................(0)

(2x+1) / ( (x-2)(x^2+1) ) = (A(x^2+1)+(Bx+C)(x-2)) / ( (x-2)(B/(x^2+1) )

(2x+1) / ( (x-2)(x^2+1) ) = (Ax^2+A+Bx^2+(C-2B)x-2C) / ( (x-2)(B/(x^2+1) )

(2x+1) / ( (x-2)(x^2+1) ) = ( (A+B)x^2+(C-2B)x+A-2C ) / ( (x-2)(B/(x^2+1) )

Match numerators

2x+1 = (A+B)x^2+(C-2B)x+A-2C

Match coefficients,

A+B = 0 ..................(1)

-2B+C = 2 .................(2)

A-2C = 1 ...................(3)

Solve for A, B and C

Substitute A from (1) in (3)

-B - 2C =1  

transpose and solve for B

B = -2C-1  ....................(4)

Substitue B from (4) in (2)

-2(-2C-1) + C = 2  

simplify

5C = 2-2 = 0

C=0  ..........................(5)

substitute (5)  in (4)

B = -2C-1 = -1  ...............(6)

Substitue (6) in (1)

A+(-1) = 0

A=1 .............................(7)

Using values from (7), (6) and (5) to substitute in (0)

we get

(2x+1) / ( (x-2)(x^2+1) ) = 1/(x-2) - x/(x^2+1)

as the required partial fraction

7 0
3 years ago
Read 2 more answers
Find the prime factorization of 2001
marusya05 [52]
Rime factorization of 2001: 

By prime factorization of 2001 we follow 5 simple steps: 
1. We write number 2001 above a 2-column table 
2. We divide 2001 by the smallest possible prime factor 
3. We write down on the left side of the table the prime factor and next number to factorize on the ride side
4. We continue to factor in this fashion (we deal with odd numbers by trying small prime factors)
5. We continue until we reach 1 on the ride side of the table

<span>2001<span>prime factorsnumber to factorize</span><span>3667</span><span>2329</span><span>291</span></span>

<span>Prime factorization of 2001 = 1×3×23×29= </span><span>1 × 3 × 23 × 29</span>
4 0
3 years ago
Complete the statement using always, sometimes, or never.
Fynjy0 [20]
I believe that, "A quadrilateral is sometimes a trapezoid”.
5 0
2 years ago
This hyperbola is centered at the
Varvara68 [4.7K]

Answer:

The equation is ( x² / 9 ) - ( y² / 7 ) = 1

Step-by-step explanation:

Given the data in question;

hyperbola is centered at the  origin, this means h and k are all equals to 0.

Vertices: (-3,0) and (3,0)

Since y-coordinates are constant, this implies it is a hyperbola with horizontal transverse axis.

h - a = -3

0 - a = -3

a = 3

Foci: (-4,0) and (4,0)

h - c = -4

0 - c = -4

c = 4

we know that, for a hyperbola

c² = a² + b²

so

⇒ ( 4 )² = ( 3 )² + b²

16 = 9 + b²

b² = 16 - 9

b² = 7

So the equation for the hyperbola will be;

⇒ ( (x-h)² / a² ) - ( (y-k)² / b² ) = 1

so we substitute

⇒ ( (x-0)² / 3² ) - ( (y-0)² / 7 ) = 1

⇒ ( x² / 3² ) - ( y² / 7 ) = 1

⇒ ( x² / 9 ) - ( y² / 7 ) = 1

Therefore, The equation is ( x² / 9 ) - ( y² / 7 ) = 1

5 0
2 years ago
Afton made a chicken dish for dinner she added a 10 ounce package of vegetables and a 14 ounces package of rice to 40 ounces of
just olya [345]
The answer your looking for is 64 pounds
3 0
3 years ago
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