All categories

4 years ago

Dy/dx when y=x^2e^5x

1 answer:

4 0

I believe the problem asks for the slope. This problem can be done by putting y=x^2e^5x into slope intercept form. However, the issue is that there is the variable "e", which does not match the slope intercept formula, y=mx+b. Long explanation short, the slope is $3e^5x^2$

You might be interested in

Which descsription means the same as this limit expression?

Nady [450]

Answer:

Step-by-step explanation:

6 0

3 years ago

X+3=44 The sum of a number and three is forty-one. Find the number.

Leto [7]

41 it’s really simple you just subtract 3 from 44 and it’s 41

7 0

3 years ago

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside one leaf of r

Anastaziya [24]

The region inside one leaf of r the area of the region is pi.20.

r =cos5θ

cos5θ =0

5θ=pi/2 ,5θ=3pi/2

θ=pi/10,θ=3pi/10

area of one leaf = ∫[pi/10 to 3pi/10](1/2)r2dθ

= ∫[pi/10 to 3pi/10](1/2)(cos5θ)2dθ

= ∫[pi/10 to 3pi/10](1/4)(1+cos10θ)dθ

=[pi/10 to 3pi/10](1/4)(θ+ (1/10)sin10θ)

=(1/4)((3pi/10)+ (1/10)sin3pi) -(1/4)((pi/10)+ (1/10)sinpi)

=(1/4)(3pi/10) -(1/4)(pi/10)

=(1/4)(2pi/10)

=pi/20

area of one leaf =pi.20.

Region area (calculus) Region area. The non-negative function given by y = f(x) represents a smooth curve on the closed interval [a, b].The area through the curve of f(x), the x-axis, and the perpendicular lines x = a and x = b The bounding region shown in Figure 1 is given by.

Learn more about the area of the region here: brainly.com/question/19053586

#SPJ4

4 0

1 year ago

In △ABC, the altitudes from vertices B and C intersect at point M, so that BM = CM. Prove that △ABC is isosceles.

AlekseyPX

Answer:

m∠MBC=m∠MCB by reason base angle theorem

Step-by-step explanation:

3 0

2 years ago

If (2i/2+i)-(3i/3+i)=a+bi, then a= A. 1/10 B. -10 C. 1/50 D. -1/10

Verizon [17]

Answer:

Option A is correct.

Step-by-step explanation:

We are given:

$\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi$

We need to find the value of a.

The LCM of (2+i) and (3+i) is (2+i)(3+i)

$=\frac{2i(3+i)}{(2+i)(3+i)}-\frac{3i(2+i)}{(2+i)(3+i)}\\=\frac{6i+2i^2}{(2+i)(3+i)}-\frac{6i+3i^2}{(2+i)(3+i)}\\=\frac{6i+2i^2-(6i+3i^2)}{(2+i)(3+i)}\\=\frac{6i+2i^2-6i-3i^2)}{5+5i}\\=\frac{-i^2}{5+5i}\\i^2=-1\\=\frac{-(-1)}{5+5i}\\=\frac{1}{5+5i}$

Now rationalize the denominator by multiplying by 5-5i/5-5i

$=\frac{1}{5+5i}*\frac{5-5i}{5-5i} \\=\frac{5-5i}{(5+5i)(5-5i)}\\=\frac{5-5i}{(5+5i)(5-5i)}\\(a+b)(a-b)= a^2-b^2\\=\frac{5(1-i)}{(5)^2-(5i)^2}\\=\frac{5(1-i)}{25+25}\\=\frac{5(1-i)}{50}\\=\frac{1-i}{10}\\=\frac{1}{10}-\frac{i}{10}$

We are given

$\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi$

Now after solving we have:

$\frac{1}{10}-\frac{i}{10}=a+bi$

So value of a = 1/10 and value of b = -1/10

So, Option A is correct.

8 0

4 years ago