? help me out please? thank you!

A radioactive substance decays exponentially with a half-life of 500 years. After 1500 years, what percent of the radioactive ma

Talja [164]

12.5%, every 500 years you take away half of the substance (50%) -> 25% -> 12.5%
you divide your number 3 times because 1500/500=3

7 0

2 years ago

F(x)=(x-6)e^-3x find the interval in which f(x) is increasing, the intervals on which f(x) is decreasing, and the local extrema

Over [174]

$\bf f(x)=(x-6)e^{-3x}\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=1\cdot e^{-3x}+(x-6)-3e^{-3x}\implies \cfrac{dy}{dx}=e^{-3x}[1-3(x-6)]
\\\\\\
\cfrac{dy}{dx}=e^{-3x}(19-3x)\implies \cfrac{dy}{dx}=\cfrac{19-3x}{e^{3x}}$

set the derivative to 0, solve for "x" to get any critical points
keep in mind, setting the denominator to 0, also gives us critical points, however, in this case, the denominator will never be 0, so... no critical points from there

there's only 1 critical point anyway, and do a first-derivative test on it, check a number before it and after it, to see what sign the derivative has, and thus, whether the graph is going up or down, to check for any extrema

5 0

3 years ago

this is formula manipulation, I'd appreciate if the steps were provided. only respond if you know how to get the answer, thank y

marissa [1.9K]

Answer:

r = 3V/(2πh²)
h = 3V/b²
r = 25/π cm ≈ 7.9577 cm
w = 15 cm

Step-by-step explanation:

1. Multiply both sides of the equation by the reciprocal of the coefficient of r.

$V\cdot\dfrac{3}{2\pi h^2}=r\\\\r=\dfrac{3V}{2\pi h^2}$

__

2. Multiply both sides of the equation by the reciprocal of the coefficient of h.

$V\cdot\dfrac{3}{b^2}=h\\\\h=\dfrac{3V}{b^2}$

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3. Solve the circumference formula for r, then substitute the given information.

$C=2\pi r\\\\r=\dfrac{C}{2\pi}\qquad\text{divide by the coefficient of r}\\\\r=\dfrac{50\,\text{cm}}{2\pi}=\dfrac{25}{\pi}\,\text{cm}\approx 7.9577\,\text{cm}$

__

4. Solve the perimeter formula for width, the substitute the given information and do the arithmetic.

$P=2(L+W)\\\\\dfrac{P}{2}=L+W\qquad\text{divide by 2}\\\\\dfrac{P}{2}-L=W\qquad\text{subtract L}\\\\\dfrac{40\,\text{cm}}{2}-5\,\text{cm}=W=15\,\text{cm}$

_____

In general, solving for a particular variable involves "undoing" what has been done to the variable, usually in the reverse order. In part 4, the variable W has L added and the sum is multiplied by 2. We "undo" those operations, last operation first, by dividing by 2 and subtracting L.

The properties of equality say you can do what you like to an equation as long as you do the same thing to both sides of the equation. So, when we say "divide by 2", we mean "divide both sides of the equation by 2." Likewise, "subtract L" means "subtract L from both sides of the equation."

4 0

3 years ago

Can someone help me with this question please?

Makovka662 [10]

Answer:

0.15 for each message

Step-by-step explanation:

The charge per message can be found by dividing the charge by the number of messages. Using the first line of the table, we get ...

charge per message = (total charge)/(number of messages) = $38.40/256

charge per message = $0.15

Alyssa is charged 0.15 for each text message.

_____

We can check other lines of the table to see if charges are really proportional:

$26.55/177 = $0.15

$31.35/209 = $0.15

6 0

3 years ago

A class of 28 students has 13 boys and 15 girls. What is the ratio of girls to boys in the class? 15:28 13:15 O 13:28 15:13 Prev

uranmaximum [27]

Answer: $\frac{15}{13}$

Step-by-step explanation:

Given

There are 28 students in a class out of which

13 are boys so the remaining 15 are girls

Ratio of girls to boys in the class $=\frac{15}{13}$

8 0

3 years ago