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2 years ago

What is the domain and range for following graph

Mathematics

1 answer:

Alex_Xolod [135]2 years ago

4 0

Answer:

the domain is -5 to 5; the range is -2.25 to 2.25

Step-by-step explanation:

the graph has no marking on it, so I suppose each block represents 1 unit

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Determine whether each expression is equivalent to 49^2t – 0.5.

vampirchik [111]

Answer:

None of the expression are equivalent to $49^{(2t - 0.5)}$

Step-by-step explanation:

Given

$49^{(2t - 0.5)}$

Required

Find its equivalents

We start by expanding the given expression

$49^{(2t - 0.5)}$

Expand 49

$(7^2)^{(2t - 0.5)}$

$7^2^{(2t - 0.5)}$

Using laws of indices: $(a^m)^n = a^{mn}$

$7^{(2*2t - 2*0.5)}$

$7^{(4t - 1)}$

This implies that; each of the following options A,B and C must be equivalent to $49^{(2t - 0.5)}$ or alternatively, $7^{(4t - 1)}$

A. $\frac{7^{2t}}{49^{0.5}}$

Using law of indices which states;

$a^{mn} = (a^m)^n$

Applying this law to the numerator; we have

$\frac{(7^{2})^{t}}{49^{0.5}}$

Expand expression in bracket

$\frac{(7 * 7)^{t}}{49^{0.5}}$

$\frac{49^{t}}{49^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{49^{t}}{49^{0.5}}$ becomes

$49^{t-0.5}}$

This is not equivalent to $49^{(2t - 0.5)}$

B. $\frac{49^{2t}}{7^{0.5}}$

Expand numerator

$\frac{(7*7)^{2t}}{7^{0.5}}$

$\frac{(7^2)^{2t}}{7^{0.5}}$

Using law of indices which states;

$(a^m)^n = a^{mn}$

Applying this law to the numerator; we have

$\frac{7^{2*2t}}{7^{0.5}}$

$\frac{7^{4t}}{7^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{7^{4t}}{7^{0.5}}$ = $7^{4t - 0.5}$

This is also not equivalent to $49^{(2t - 0.5)}$

C. $7^{2t}\ *\ 49^{0.5}$

$7^{2t}\ *\ (7^2)^{0.5}$

$7^{2t}\ *\ 7^{2*0.5}$

$7^{2t}\ *\ 7^{1}$

Using law of indices which states;

$a^m*a^n = a^{m+n}$

$7^{2t+ 1}$

This is also not equivalent to $49^{(2t - 0.5)}$

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Answer:

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