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3 years ago

Is the following true or false? d/dx[x^3e^x]=x^2e^x (3x+2)

1 answer:

3 0

It's false. It's a product so...
Derivative of the first TIMES the second PLUS derivative of second TIMES the first.

Derivative of the first (x^3) = 3x^2
Times the second = 3x^2 * e^x

Derivative of the second = e^x (remains unchanged)
Times the first = e^x * x^3
So the answer would be (3x^2)(e^x) + (e^x)(x^3)
which can be factorised to form x^2·e^x(3 + x)

You might be interested in

PLEASE HELP! WILL MARK BRAINLIEST!!

adoni [48]

Answer:

41.04 meters

Step-by-step explanation:

The questions which involve calculating the angles and the sides of a triangle either require the sine rule or the cosine rule. In this question, the two sides that are given are adjacent to each other the given angle is the included angle. The initial position is given by A. The tree is denoted as C and the fence post is denoted as B. Since the use of sine rule will complicate the question, it will be easier to solve this question using the cosine rule. Therefore, cosine rule will be used to calculate the length of BC. The cosine rule is:

BC^2 = AB^2 + AC^2 - 2*AB*AC*cos(BAC).

The question specifies that AC = 70 meters, BAC = 25°, and AB = 35 meters. Plugging in the values:

BC^2 = 35^2 + 70^2 - 2(35)(70)*cos(25°).

Simplifying gives:

BC^2 = 1684.091844.

Taking square root on the both sides gives BC = 41.04 meters (rounded to two decimal places).

Therefore, the distance between the point on the tree to the point on the fence post is 41.04 meters!!!

6 0

3 years ago

Suppose M is the midpoint of line FG.

alukav5142 [94]

$A)\\|FM|=5y+13;\ |MG|=5-3y\\\\|FG|=|FM|+|MG|\\\\|FG|=(5y+13)+(5-3y)=2y+18\\\\B)\\|FM|=8a+1;\ |FG|=42\\\\|FG|=2|FM|\\\\2(8a+1)=42\\16a+2=42\ \ \ \ \ |subtract\ 2\ from\ both\ sides\\16=40\ \ \ \ \ |divide\ both\ sides\ by\ 16\\a=\frac{40}{16}\\a=\frac{5}{2}\\a=2.5$

6 0

3 years ago

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ

bulgar [2K]

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

8 0

3 years ago

A taco stand sells tacos for $3.25 each. The stands expenses for the day are $210. Define your variables and write an inequality

Julli [10]

Given:

Let P= profit
let n= no. of tacos sold per day

Sol'n:

P= 3.25n-210
the profit needs to be positive, thus

3.25n>210

n>210/3.25

n> 64.615

they can only sell whole tacos, therefore they must sell at least 65 tacos to make a profit and that profit is:

P=3.25n-210

P=3.25*65 - 210

P= 211.25 -210

P= 1.25

$1.25 profit

3 0

3 years ago

A bag contains blue marbles and red marbles, 56 in total. The number of blue marbles is 8 more than 2 times the number of red ma

Nikolay [14]

24 Blue marbles.

Explanation:

56 - 8 = 48

48 / 2 = 24

so if there are 24 blue marbles, 24 x 2 = 48, and then 8 more is 56.

8 0

3 years ago

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