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Ipatiy [6.2K]
3 years ago
9

What is the value of x? Leave your answer in simplest radical form.

Mathematics
2 answers:
Alja [10]3 years ago
8 0

Answer:

Step-by-step explanation:

As it is right angle triangle, we can use Pythagoras theorem

x² = 4² + 6²

x² = 16 + 36

x² = 52

x = √52 = √2*2*13

x = 2√13

OleMash [197]3 years ago
5 0

Step-by-step explanation:

By using Pythagoras theoram,

x²= (6)²+(4)²

{x}^{2}  = 36 + 16

{x}^{2}  = 52

x =  \sqrt{52}

x = 2 \sqrt{13}

optionD

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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)

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Answer:

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