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Zina [86]
1 year ago
11

Simplify the expression below. 2^3 x 2^2 A.46 B. 45 C. 25 D. 26

Mathematics
2 answers:
bixtya [17]1 year ago
4 0

\huge\text{Hey there!}


\huge\boxed{E}\textsf{quation:}

\mathsf{2^3 \times 2^2}


\huge\boxed{S}\textsf{olving:}

\mathsf{2^3 \times 2^2}

\mathsf{= 2\times2\times2 \times2\times2}

\mathsf{= 4\times4\times2}

\mathsf{= 16\times2}

\mathsf{= 32}


\huge\boxed{T}\textsf{herefore, your answer should be:}

\huge\boxed{\frak{32}}\huge\checkmark

\huge\text{Good luck on your assignment \& enjoy your day!}


~\frak{Amphitrite1040:)}

Maru [420]1 year ago
3 0

Answer: 32 is the answer to the expression you provided.

Step-by-step explanation:

2^3\times2^2\\2\times2\times2\times2\times2\\4\times4\times2\\16\times2\\\rightarrow32

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The value of a certain car decreases by 16% each year. What is the 1⁄2-life of the car?
svet-max [94.6K]

Answer:

The half life of the car is 3.98 years.

Step-by-step explanation:

The value of the car after t years is given by the following equation:

V(t) = V(0)(1-r)^{t}

In which V(0) is the initial value and r is the constant decay rate, as a decimal.

The value of a certain car decreases by 16% each year.

This means that r = 0.16

So

V(t) = V(0)(1-r)^{t}

V(t) = V(0)(1-0.16)^{t}

V(t) = V(0)(0.84)^{t}

What is the 1⁄2-life of the car?

This is t for which V(t) = 0.5V(0). So

V(t) = V(0)(0.84)^{t}

0.5V(0) = V(0)(0.84)^{t}

(0.84)^{t} = 0.5

\log{(0.84)^{t}} = \log{0.5}

t\log{0.84} = \log{0.5}

t = \frac{\log{0.5}}{\log{0.84}}

t = 3.98

The half life of the car is 3.98 years.

7 0
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