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Arada [10]
3 years ago
8

e=" \frac{2}{3} - \frac{4}{10} = " alt=" \frac{2}{3} - \frac{4}{10} = " align="absmiddle" class="latex-formula">
what is the answer?
Mathematics
1 answer:
rusak2 [61]3 years ago
6 0

Answer:

8/30 (Simplified answer: 4/15)

Step-by-step explanation:

Find common denominators. Note the what you do to the denominator, you do to the numerator.

Multiply 10 to both the denominator & numerator of 2/3, and multiply 3 to both the denominator & numerator of 4/10

(10/10)(2/3) - (3/3)(4/10)

Simplify.

(20/30) - (12/30)

Simplify. Combine the terms

(20/30) - (12/30) = (20 - 12)/30 = 8/30

Simplify the fraction

(8/30)/(2/2) = 4/15

~

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Find the value of the expression (x2−4)3x when x = 5.
Degger [83]

Answer:

315

Step-by-step explanation:

You plug 5 in for x

(5^2-4)3(5) = 315

7 0
2 years ago
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Show that a = −1 + √3i and b = 2 satisfy 1/a+b=1/a + 1/b
Zarrin [17]

Answer:

LHS = \frac{1 - \sqrt3i}{4} = RHS = \frac{1 - \sqrt3i}{4}

Step-by-step explanation:

Data provided in the question:

a = −1 + √3i and b = 2

to prove:

\frac{1}{a+b}=\frac{1}{a} + \frac{1}{b}

Considering the LHS

⇒ \frac{1}{a+b}

substituting the value of a and b, we get

⇒ \frac{1}{−1 + \sqrt3i+2}

or

⇒ \frac{1}{1 + \sqrt3i}

on multiplying and dividing by conjugate ( 1 - √3i )

we get

\frac{1}{1 + \sqrt3i}\times\frac{1 - \sqrt3i}{1 - \sqrt3i}

or

\frac{1 - \sqrt3i}{(1^2 - (\sqrt3i)^2}

or

\frac{1 - \sqrt3i}{1 + 3}              (as (√i)² = -1 )

or

\frac{1 - \sqrt3i}{4}

Now,

considering the RHS

\frac{1}{a} + \frac{1}{b}

substituting the value of a and b, we get

⇒ \frac{1}{-1 + \sqrt3i} + \frac{1}{2}

or

⇒ \frac{2\times1 + ( -1 + \sqrt3i)\times1}{(-1 + \sqrt3i)\times2}

or

⇒ \frac{2 + ( -1 + \sqrt3i)}{(-1 + \sqrt3i)\times2}

or

⇒ \frac{1 + \sqrt3i}{(-1 + \sqrt3i)\times2}

now,

on multiplying and dividing by conjugate ( -1 - √3i )

we get

\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1 - \sqrt3i}{-1 - \sqrt3i}

or

\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1( 1 + \sqrt3i)}{-1 - \sqrt3i}

or

\frac{(1 + \sqrt3i}^2\times(-1){((-1)^2 - (\sqrt3i)^2)\times2}

or

\frac{(1^2 + (\sqrt3i)^2+2(1)(\sqrt3i)\times(-1)}{(1 + 3)\times2}

or

\frac{(1 - 3 + 2\sqrt3i)\times(-1)}{(4)\times2}

or

\frac{(-2 + 2\sqrt3i)\times(-1)}{(4)\times2}

or

\frac{-2( 1 - 2\sqrt3i)\times(-1)}{(4)\times2}

or

\frac{( 1 - 2\sqrt3i)}{(4)}

Since, LHS = RHS

hence satisfied

3 0
3 years ago
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Hello there. Sorry from last time.

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<span>B.12.55</span>


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Please give answer quickly in simplest form
abruzzese [7]
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hope i helped!

7 0
3 years ago
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KiRa [710]

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