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3 years ago

Find the value of N please!!

Mathematics

2 answers:

V125BC [204]3 years ago

6 0

Answer:

n = - 5

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ ⇔ $a^{(m+n)}$

$\frac{a^{m} }{a^{n} }$ ⇔ $a^{(m-n)}$

Consider the left side

$\frac{y^4.y^{n} }{y^2}$

= $\frac{y^{4+n} }{y^2}$

= $y^{4+n-2}$

= $y^{2+n}$ , then

$y^{2+n}$ = $y^{-3}$

Since the bases on both sides are the same, equate the exponents

2 + n = - 3 ( subtract 2 from both sides )

n = - 5

babymother [125]3 years ago

6 0

What my mans above said

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4 years ago

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aniked [119]

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4 years ago

Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants. 5x − 4 x(x2 + 7

Nikitich [7]

Answer:

$\frac{5x-4}{x(x^2+7)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+7} + \frac{Dx+E}{(x^2+7)^2}$

Step-by-step explanation:

Given the expression $\frac{5x-4}{x(x^2+7)^2}$ , we are to re-write the expression in form of a partial fraction.

Before we write in form of a partial fraction, we need to note the expression at the denominator. Since the expression in parenthesis is a quadratic equation, the equivalent numerator must be a linear expression.

Also the quadratic equation is a repeated form since it is squared. This means that we are to repeat the quadratic equation twice when writing as a partial fraction.

$\frac{5x-4}{x(x^2+7)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+7} + \frac{Dx+E}{(x^2+7)^2}$

From the above partial fraction, it can be seen that x² + 7 in parenthesis was repeated twice and their equivalent expressions at the numerator are both linear i.e Bx+E and Dx+ E where A, B, C, D and E are the unknown constant.

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3 years ago

Solve the equation on the

Kryger [21]

Answer:

see explanation

Step-by-step explanation:

note that (sinx)² = sin²x

4sin²x - 2 = 0 ( add 2 to both sides )

4sin²x = 2 ( divide both sides by 4 )

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