A market buys cashews for $15 per pound. They want to mark up the cashews 20%. What is the new price per pound?

A certain compound has a half-life of four days. Write and use an exponential decay function to find the amount of compound rema

dimaraw [331]

Answer:

1.97 ounces

Step-by-step explanation:

-The exponential decay function is given by the formula:

$P_t=P_oe^{-rt}\\\\$

Given that the half-life is 4 days, we can substitute in the formula to solve the rate of decay for a single half-life:

$P_t=P_oe^{rt}\\\\P_t=0.5P_o\\\\\therefore 0.5=e^{-4r}\\\\-4r=In(0.5)\\\\r=0.17329$

#Since, 1 week has 7 days, 3 weeks is a 21-day period, thus, t=21.

#We substitute and solve for the amount remaining after 21days of decay:

$P_t=P_oe^{-rt}\\\\=75e^{-21\times 0.17329}\\\\=1.97072\\\\\approx1.97 \ oz$

Hence, the amount remaining after 3 weeks of decays is approximately 1.97 ounces

4 0

3 years ago

Elsa is going to rent a truck for one day. There are two companies she can choose from. Company A charge $85 and allows unlimite

Gnom [1K]

Elsa is going to buy a truck so she can give down to the good old bar to drink her painful sadness away so that later she'll feel happy enough to take the barrel of gun and put it in her mouth and just starts deepthoarting

5 0

3 years ago

Read 2 more answers

On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes

belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

$r = \frac{\Delta T_{X}}{\Delta T_{Y}}$

$r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}$

$r = 11\,\frac{^{\circ}X}{^{\circ}Y}$

The difference between current temperature in Y linear scale with respect to freezing point is:

$\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)$

$\Delta T_{Y} = 115\,^{\circ}Y$

The change in X linear scale is:

$\Delta T_{X} = r\cdot \Delta T_{Y}$

$\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)$

$\Delta T_{X} = 1265\,^{\circ}X$

Lastly, the current temperature on the X scale is:

$T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X$

$T_{X} = 1150\,^{\circ}X$

The current temperature on the X scale is 1150 °X.

5 0

3 years ago

Find missing angles

yKpoI14uk [10]

Hey

The value of X is 35°

X and 35° are vertically opposite angles.

Vertically opposite angles are always equal.

Hope it helps

Good luck on your assignment

8 0

3 years ago

Which of the following have the property that a(x)=a−1(x)? I. y=x II. y=1/x III.y=x^2 IV. y=x^3 A. I and II, only B. IV, only C.

valentina_108 [34]

Answer:

Correct answer:

A. I and II

Step-by-step explanation:

First of all, let us have a look at the steps of finding inverse of a function.

1. Replace y with x and x with y.

2. Solve for y.

3. Replace y with $f^{-1}(x)$

Given that:

$I.\ y=x \\II.\ y=\dfrac{1}x \\III.\ y=x^2 \\IV.\ y=x^3$

Now, let us find inverse of each option one by one.

I. y = x, a(x) = x

Replacing y with and x with y:

x = y

x = $a^{-1}(x)$ = $a(x)$ Hence, I is true.

II. $y =\dfrac{1}{x}$

Replacing y with and x with y:

$x =\dfrac{1}{y}$

$x=\dfrac{1}{a^{-1}(x)}$

$\Rightarrow a^{-1}(x) = \dfrac{1}{x}$

$a^{-1}(x)$ = $a(x)$ Hence, II is true.

III. $y =x^{2}$

Replacing y with and x with y:

$x =y^{2}\\\Rightarrow y = \sqrt x\\\Rightarrow a^{-1}(x) = \sqrt{x} \ne a(x)$

Hence, III is not true.

IV. $y =x^{3}$

Replacing y with and x with y:

$x =y^{3}\\\Rightarrow y = \sqrt[3] x\\\Rightarrow a^{-1}(x) = \sqrt[3]{x} \ne a(x)$

Hence, IV is not true.

Correct answer:

A. I and II

4 0

3 years ago