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My name is Ann [436]
3 years ago
5

Please hepl with math

Mathematics
1 answer:
Ivenika [448]3 years ago
6 0

Answer:

\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}} = -\dfrac{5}{3}

Step-by-step explanation:

We are given the expression to be simplified:

\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}}

Let us take common a term with a power of 5 from the numerator and the denominator of the given expression.

We know that:

a^{p+q} = a^p \times a^q

Let us use it to solve the powers of 5 in the given expression.

\therefore we can write:

5^{n+2} = 5^{n+1}\times 5= 5^n\times 5^{2}

5^{n+1} = 5^n\times 5

The given expression becomes:

\dfrac{5^{n+1} \times 5-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n}\times 5}

Taking common 5^{n+1} from the numerator and

Taking common 5^{n} from the denominator

\Rightarrow \dfrac{5^{n+1} (5-6)} {5^{n}(13-2\times5)}\\\Rightarrow \dfrac{5^{n+1} (-1)} {5^{n}(13-10)}\\\Rightarrow -\dfrac{5^{n+1}} {5^{n}\times3}\\\Rightarrow -\dfrac{5^{n}\times 5} {5^{n}\times3}\\\Rightarrow -\dfrac{5}{3}

\therefore The answer is:

\dfrac{5^{n+2}-6\times 5^{n+1}}{13 \times5^{n}-2\times5^{n+1}} = -\dfrac{5}{3}

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Answer:

\frac{c}{a} and \frac{b}{a}

Step-by-step explanation:

sinB = \frac{opposite}{hypotenuse} = \frac{AC}{BC} = \frac{b}{a}

tanC = \frac{opposite}{adjacent} = \frac{AB}{AC} = \frac{c}{b}

Thus

sinB tanC = \frac{b}{a} × \frac{c}{b} ( cancel b on numerator/ denominator )

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---------------------------------------------------------------------------

sinC = \frac{opposite}{hypotenuse} = \frac{AB}{BC} = \frac{c}{a}

tanB = \frac{opposite}{adjacent} = \frac{AC}{AB} = \frac{b}{c}

Thus

sinC tanB = \frac{c}{a} × \frac{b}{c} ( cancel c on numerator/ denominator )

                 = \frac{b}{a}

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3 years ago
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To find the answer, you would do 6 divided by 4.2</span>
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Step-by-step explanation:

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