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allsm [11]
3 years ago
13

A rectangle and a chai go have the same area. If their bases are the same lengths, how do their Heights compare? Justify your an

swer.
Mathematics
1 answer:
Alika [10]3 years ago
7 0
I think u mean triangle in your question.
Area of rectangle = l × b = height × base = bh
Area of traingle = ½bh'
Area of rectangle = Area of triangle (given)
bh = ½bh'
h' = 2h
It means that the height of the given triangle is double the height of the rectangle.
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Answer:

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Step-by-step explanation:

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Which shows the following expression using positive exponents?
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Answer:

1. option B is correct i.e., \frac{-3a^{7}b^3b^1}{15a^{2}}

2. option D is correct i.e., \frac{a^3b^4}{ab^2}.

3. option B is correct i.e.,  \frac{n^{10}}{16m^{6}}.

Step-by-step explanation:

1. Since the given equation is:

\frac{-3a^{-2}b^3}{15a^{-7}b^{-1}}

As we know that a^{-x}=\frac{1}{a^x}

So, the given equation can be represented as:

\frac{-3a^{7}b^3b^1}{15a^{2}}


2. Given equation is:

\frac{a^3b^{-2}}{ab^{-4}}

As we know that a^{-x}=\frac{1}{a^x}

so the given equation can be represented as:

\frac{a^3b^4}{ab^2}.


3.  Given equation is:

(\frac{4mn}{m^{-2}n^6})^{-2}

by using the properties a^{-x}=\frac{1}{a^x},

and a^x\times a^y=a^{x+y}

(\frac{4mn}{m^{-2}n^6})^{-2}

= (\frac{4n^1n^{-6}}{m^{-2}m^{-1}})^{-2}

= (\frac{4n^{1+(-6)}}{m^{-2+(-1)}})^{-2}

= (\frac{4n^{-5}}{m^{-3}})^{-2}

= \frac{4^{-2}n^{-5\times (-2)}}{m^{-3\times (-2)}}

= \frac{n^{10}}{16m^{6}}.

which is option B.

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