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3 years ago

Choose the expression that completes the equation.

Mathematics

2 answers:

In-s [12.5K]3 years ago

6 0

Answer:

Option 3

Step-by-step explanation:

33 / 3 + 10 = 21

1.) 5(3 + 5) = 40

2.) 10(1 + 1) = 22

3.) 3(6 + 1) = 21

Harrizon [31]3 years ago

5 0

Answer:

Answer would be 3

Step-by-step explanation:

2(10+1)= 21 and

33 divided by 3 + 10 = 21

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Which expressions are equivalent to \dfrac{4^{-3}}{4^{-1}} 4 −1 4 −3 start fraction, 4, start superscript, minus, 3, end super

Slav-nsk [51]

Answer:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}$

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}$

Step-by-step explanation:

Given

$\dfrac{4^{-3}}{4^{-1}}$

Required

Choose equivalent expressions

Choosing the first answer:

$\dfrac{4^{-3}}{4^{-1}}$

Split expressions

$4^{-3} * \frac{1}{4^{-1}}$

Apply laws of indices: $(a^{-b} = \frac{1}{a^b})$

$\frac{1}{4^3} * \frac{1}{4^{-1}}$

Apply laws of indices: $(a^{-b} = \frac{1}{a^b})$

$\frac{1}{4^3} * \frac{1}{1/4}$

$\frac{1}{4^3} * \frac{4^1}{1}$

$\frac{4^1}{4^3}$

Hence:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}$

Choosing the second:

$\dfrac{4^{-3}}{4^{-1}}$

Apply law of indices: $(\frac{a^m}{a^n} = a^{m-n})$

So,

$\dfrac{4^{-3}}{4^{-1}} = 4^{-3-(-1)}$

$\dfrac{4^{-3}}{4^{-1}} = 4^{-3+1)}$

$\dfrac{4^{-3}}{4^{-1}} = 4^{-2}$

Apply law of indices: $(a^{-b} = \frac{1}{a^b})$

So:

$\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}$

4 0

3 years ago

What is 500 divided by 20

Nikolay [14]

The answer is 25 just use a calculator for that type of math

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Find the domain of each function. (Enter your answers using interval notation.)(a) f(x)=8/(1+e^x) .(b) f(x)=5/(1−e^x)

Alexus [3.1K]

Answer:

domain is (-∞,∞)

(-∞,0) U(0,∞)

Step-by-step explanation:

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In f(x) there is no restriction for x. the function is defined for all x values

so domain is (-∞,∞)

$f(x)=\frac{5}{1-e^x}$

The domain is the set of x values for which the function is defined

when x=0, e^0=1 that makes the denominator 0

Denominator 0 is undefined. So x cannot be 0

domain is (-∞,0) U(0,∞)

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