Find the derivative of e^(lnx)^2

help me plz you'll get brainliest if u show work and the only way for u to help me is if u show your work so plz do

Usimov [2.4K]

Answer:

D.

Step-by-step explanation:

This will be D because the beaker already has 0.2 mL of water. If Jovita needs to add more water to the beaker, and then it is 0.8. In D, the diagram shows you 0.2 + x = 0.8, which makes it the right diagram.

Hope that helps!

5 0

2 years ago

Read 2 more answers

The price of products may increase due to inflation and decrease due to depreciation. Marco is studying the change in the price

likoan [24]

Answer:

A) 3%

B) Product A

Step-by-step explanation:

Exponential Function

General form of an exponential function: $y=ab^x$

where:

a is the initial value (y-intercept)
b is the base (growth/decay factor) in decimal form
x is the independent variable
y is the dependent variable

If b > 1 then it is an increasing function

If 0 < b < 1 then it is a decreasing function

Part A

Product A

Assuming the function for Product A is exponential:

$f(x) = 0.69(1.03)^x$

The base (b) is 1.03. As b > 1 then it is an increasing function.

To calculate the percentage increase/decrease, subtract 1 from the base:

⇒ 1.03 - 1 = 0.03 = 3%

Therefore, product A is increasing by 3% each year.

Part B

$\sf percentage\:change=\dfrac{final\:value-initial\:value}{initial\:value} \times 100$

To calculate the percentage change in Product B, use the percentage change formula with two consecutive values of f(t) from the given table:

$\implies \sf percentage\:change=\dfrac{10201-10100}{10100}\times 100=1\%$

Check using different two consecutive values of f(t):

$\implies \sf percentage\:change=\dfrac{10303.01-10201}{10201}\times 100=1\%$

Therefore, as 3% > 1%, Product A recorded a greater percentage change in price over the previous year.

Although the question has not asked, we can use the given information to easily create an exponential function for Product B.

Given:

a = 10,100
b = 1.01
n = t - 1 (as the initial value is for t = 1 not t = 0)

$\implies f(t) = 10100(1.01)^{t-1}$

To check this, substitute the values of t for 1 through 4 into the found function:

$\implies f(1) = 10100(1.01)^{1-1}=10100$

$\implies f(2) = 10100(1.01)^{2-1}=10201$

$\implies f(3) = 10100(1.01)^{3-1}=10303.01$

$\implies f(4) = 10100(1.01)^{4-1}=10406.04$

As these values match the values in the given table, this confirms that the found function for Product B is correct and that Product B increases by 1% per year.

4 0

2 years ago

Please help me with this one

lara31 [8.8K]

Answer:

D

Step-by-step explanation:

D, it has a right angle and it shows us that the hypotenuse is congruent and the leg is also congruent.

7 0

3 years ago

A circular pool is 3.5 meters deep and has a radius of 9 meters. What is the volume

mote1985 [20]

Answer:

$890.75m^3$

Step-by-step explanation:

Step one:

given data

heigth/depth = 3.5m

radius= 9m

Step two:

The shape is that of a cylinder

Hence the volume of the pool is

$V= \pi r^2h$

substitute

$V= 3.142*9^2*3.5\\\\V=890.75 m^3$

7 0

3 years ago

A poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ag

Katyanochek1 [597]

Answer:

At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll

Step-by-step explanation:

Given that a poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ago.

Let five years ago be group I X and as of now be group II Y

$H_0: p_x =p_y\\H_a: p_x \neq p_y$

(Two tailed test at 5% level of significance)

Group I Group II combined p

n 270 300 570

favor 120 140 260

p 0.4444 0.4667 0.4561

Std error for differene = $\sqrt{\frac{0.4561(1-0.4561)}{570} } \\=0.0209$

p difference = -0.0223

Z statistic = p diff/std error = -1.066

p value =0.2864

Since p value >0.05, we accept null hypothesis.

At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll

7 0

3 years ago