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3 years ago

The answer to this...3×25/6

1 answer:

6 0

Ur answer is
12.5
heres what helps me when im stuck hope it helps u =^-^=
pemdas steps
parentheses 1st
exponent 2nd
multiply 3rd
divide 4th
add 5th
subtract 6th

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0.08 is 10 times as great

Vlad [161]

Answer: 0.08 is 10 times as great as 0.008

Step-by-step explanation: 0.008 x 10 = 0.08

7 0

3 years ago

Read 2 more answers

At the Cubs/White Sox baseball game on Sunday night, 65% of the fans were men. Another 25% were women. Of the remainder, 40% wer

strojnjashka [21]

Ok, 65% of the attendants were men, 25% were women.

That leaves us with 10% remaining.

From those 10% .................The problem tells us that 40% are boys

So 10% * ( 0.4 ) = 4% ..............This means that 4% of the total attendants were boys.

So the distribution is

65%........men
25%........women
4%..........boys
6%...........girls

640 boys attended the game.....> this means that 4% of the total amount of people is 640.

So the number of persons that attended were

T= 100% *(640)) / 4% = 16000 people

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3 years ago

3-8 Present Value of Investments 165

Viefleur [7K]

It's so easy just keep on reading it

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3 years ago

Find the quotient. Simplify if possible. <br><br> What is 2/9 divided by (-3/27) ?

Black_prince [1.1K]

$\sf \frac{2}{9}:(-\frac{3}{27}^{(3})= \frac{2}{9}* (-9)=\boxed{\sf -2}$

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4 years ago

I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.