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tamaranim1 [39]
3 years ago
6

Use a number line to show the sum of 6 + (–7).

Mathematics
1 answer:
mylen [45]3 years ago
4 0
So there is a rule
(+)(+)=+
(-)(+)= -
(+)(-)= -
So now if you subtract 6 with 7 you will end up having -1 as you’re answer. Hope this helps
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What is the percent decrease from 60 to 39?
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X - the percent decrease

60-60x=39 \ \ \ |-60 \\&#10;-60x=-21 \ \ \ |\div (-60) \\&#10;x=\frac{-21}{-60} \\ \Downarrow \\ x=\frac{-21}{-60}=\frac{-21 \div (-3)}{-60 \div (-3)}=\frac{7}{20}=\frac{7 \times 5}{20 \times 5}=\frac{35}{100}=35\%

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the third one

Step-by-step explanation:

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2 years ago
Please hepl with math
Ivenika [448]

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:

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