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2 years ago

Which of the following shows the prime factorization of 36 using exponential notation?

Mathematics

1 answer:

Daniel [21]2 years ago

4 0

Answer:

The prime factorization of 36 using exponential notation is $2^2 \times 3^2$

Step-by-step explanation:

Given: number 36

We have to show the prime factorization of 36 using exponential notation.

Prime factorization is representing a number in form of products of its primes.

Consider the given number 36

36 can be written as 2 × 2 × 3 × 3

In exponential notation it can be written as $2^2 \times 3^2$

Thus, the prime factorization of 36 using exponential notation is $2^2 \times 3^2$

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The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from th

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$\bf \qquad \textit{ inverse proportional variation}\\\\\\
\begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
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"The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth"

namely "w" varies inversely to $\bf d^2$ or $\bf w=\cfrac{k}{d^2}$

then
"A particular person weighs 192 pounds on the surface of the earth and the radius of the earth is 3900 miles. "
namely when w = 192, d = 3900

so $\bf w=\cfrac{k}{d^2}\qquad 
\begin{cases}
w=192\\
d=3900
\end{cases}\implies 192=\cfrac{k}{3900^2}
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2 years ago

Research reports indicate that surveillance cameras at major intersections dramatically reduce the number of drivers who barrel

Tju [1.3M]

Answer:

(a)Increasing

(b)t=1.34 years

Step-by-step explanation:

Given the function

N(t) = 5.85t³-23.43t²+45.06t+69.5, 0≤t≤4

(a)N(0)=5.85(0)³-23.43(0)²+45.06(0)+69.5=69.5

N(4)=5.85(4)³-23.43(4)²+45.06(4)+69.5=249.26

A function is increasing whenever x₁≤x₂, f(x₁)≤f(x₂).

Since in the interval (0,4), N(0)<N(4), we say the function is increasing.

(b)The number of communities using surveillance cameras at intersections changed least rapidly at the point where the derivative of the function is zero.

N(t) = 5.85t³-23.43t²+45.06t+69.5

N'(t)=17.49t²-46.86t+45.06

If N'(t)=0,

17.49t²-46.86t+45.06=0

Solving the quadratic equation gives the values of t as:

t=1.3396-0.8842i

t=1.3396+0.8842i

We take the Real Part as our Minimum value,

The time when number of communities using surveillance cameras at intersections changed least rapidly is:

t=1.34(to 2 decimal places)

(c)Rate of Increase using a security camera/year.

N'(t)=17.49t²-46.86t+45.06

N'(t)=17.49(1)²-46.86(1)+45.06

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2 years ago

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