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skad [1K]
3 years ago
5

Help can't figure this problem out

Mathematics
1 answer:
Darya [45]3 years ago
8 0
(x^1/2 y^-2/3)^-6
= x^(1/2*-6) y^[(-2/3) *(-6)]
= x^-3 y ^4
= y^4 / x^3

answer is D.  y^4 / x^3  
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Determine if the function f(x)=x+4 is increasing or decreasing on the interval of 2≤x≤5. Justify your answer.
Rudik [331]

Step-by-step explanation:

Given

The function is f(x)=x+4

for x_2>x_1, we will check the values of f(x_1) and f(x_2)

Take 3 and 4

f(4)=4+4\\f(4)=8

Similarly,

f(3)=3+4\\f(3)=7

Thus, f(4)>f(3)

Also, f'(x) is 1 i.e. a constant value. So, the function is increasing linearly for the given interval.

7 0
3 years ago
If a rectangle has a perimeter of 58 ft and a length of 4 feet what is the width of the rectangle?
faltersainse [42]

Answer:

25ft

Hoped this helped!

5 0
3 years ago
Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =
marta [7]

Answer:

The integral for the arc of length is:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx

By using Simpon’s rule we get: 1.5355453

And using technology we get:  2.3020

The approximation is about 33% smaller than the exact result.

Explanation:

The formula for the length of arc of the function f(x) in the interval [a,b] is:

\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx

We need the derivative of the function:

f'(x)=\frac{1}{x}

And we need it squared:

[f'(x)]^2=\frac{1}{x^2}

Then the integral is:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx

Now, the Simposn’s rule with n=4 is:

\displaystyle\int_a^b g(x)}dx\approx\frac{\Delta x}{3}\left( g(a)+4g(a+\Delta x)+2g(a+2\Delta x) +4g(a+3\Delta x)+g(b) \right)

In this problem:

a=1,b=3,n=4, \displaystyle\Delta x=\frac{b-a}{n}=\frac{2}{4}=\frac{1}{2},g(x)= \sqrt{1+\frac{1}{x^2}}

So, the Simposn’s rule formula becomes:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)

Then simplifying a bit:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)

Then we just do those computations and we finally get the approximation via Simposn's rule:

\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\approx 1.5355453

While when we do the integral by using technology we get: 2.3020.

The approximation with Simpon’s rule is close but about 33% smaller:

\displaystyle\frac{2.3020-1.5355453}{2.3020}\cdot100\%\approx 33\%

8 0
3 years ago
Which is true given the two equations below?
yKpoI14uk [10]
A because the intersection point of the graph is 3x-8y=19
6 0
3 years ago
Write an equation of the line that passes through (2,−5) and is parallel to the line 2y=3x+10.
r-ruslan [8.4K]

Step-by-step explanation:

the eqn is 3x-2y+10=0

slope m=-3/-2=3/2 (m=(-xcoeff/ycoeff)

eqn is

y-y1=m(x-x1)

y+5=3/2(x-2)

2y+10=3x-6

3x-2y-16=0

8 0
3 years ago
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