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3 years ago

Is 3^9 greater than 7^5? How much less than or greater than is 3^9 compared to 7^5? Show your work

Mathematics

1 answer:

ki77a [65]3 years ago

5 0

Is 3^9 greater than 7^5?

$3^9 = 3*3*3*3*3*3*3*3*3= 19683$

$7^5 = 7*7*7*7*7=16807$

19683 is greater than 16807

so $3^9 \ greater \ than \ 7^5$

Now we compare $3^9 \ and \ 7^5$

for that we compare 19683 and 16807

19683 - 16807= 2876

19683 is 2876 more than 16807

So , $3^9$ is 2876 greater than $7^5$

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The function is $P(x) = \frac{1}{2} e^{-x^2} +0.016$

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From the question we are told that

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$P'(x) = xe^{-x^2}$ is the rate of growth

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$P(x) = \int\limits {P'(x)} \, = \int\limits {xe^{-x^2}} \,$

So using substitution method

We have that

$u = - x^2$

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$p(x) = \int\limits {\frac{1}{2} e^{-u}} \, du$

$p(x) = {\frac{1}{2} [ e^{-u}} +c ]$

$p(x) = {\frac{1}{2} e^{-x^2}} + \frac{1}{2} c$ recall $u = - x^2$ and let $\frac{1}{2} c = Z$

At x = 2 years

$P(x) =$ $25,000

Since the profit rate is in million

$P(x) =$ $25,000 = $\frac{25000}{1000000} =$ $0.025 millon dollars

$0.025 = {\frac{1}{2} e^{-2^2}} + Z$

=> $Z = 0.025 - {\frac{1}{2} e^{-2^2}}$

$Z = 0.016$

So the profit function becomes

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3 years ago

NEED HELP ASAP

VARVARA [1.3K]

When solving an equation with an absolute value term, you make two separate equations ans solve for x:

Equation 1: |4x-3|-5 = 4

1st add 5 to both sides:

|4x-3| = 9

Remove the absolute value term and make two equations:

4x-3 = 9 and 4x - 3 = -9

Solving for x you get X = 3 and x = -1.5

When you replace x with those values in the original equation the statement is true so those are two solutions.

Do the same thing for equation 2:

|2x+3| +8 = 3

Subtract 8 from both sides:

|2x+3| = -5

Remove the absolute value term and make two equations:

2x +3 = -5

2x+3 = 5

Solving for x you get -1 and 4, but when you replace x in the original equation with those values, the statement is false, so there are no solutions.

The answer is:

C. The solutions to equation 1 are x = 3, −1.5, and equation 2 has no solution.

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