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2 years ago

Find the value of y.

Mathematics

1 answer:

malfutka [58]2 years ago

4 0

Angle at the circumference is half the angle at the centre of the circle subtended by the same arc
y=65

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There are 6 math classes in 8th grade. 5 are equally sized and the 6th class has 32 students. If there are a total of 202 studen

butalik [34]

Answer:

34+34+34+34+34=202 there u go!

5 0

1 year ago

Complete the recursive formula of the arithmetic sequence -3, -1, 1, 3,...

Strike441 [17]

Given:

The given arithmetic sequence is:

$-3,-1,1,3,...$

To find:

The recursive formula of the given arithmetic sequence.

Solution:

We have,

$-3,-1,1,3,...$

Here, the first term is -3. So, $b(1)=-3$ .

The common difference is:

$d=-1-(-3)$

$d=-1+3$

$d=2$

The recursive formula of an arithmetic sequence is:

$b(n)=b(n-1)+d$

Where, d is the common difference.

Putting $d=2$ , we get

$b(n)=b(n-1)+2$

Therefore, the recursive formula of the given arithmetic sequence is $b(n)=b(n-1)+2$ , where $b(1)=-3$ .

8 0

3 years ago

Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p

Juli2301 [7.4K]

I'm reading this as

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy$

with $\nabla f=(2xe^{-y},2y-x^2e^{-y})$ .

The value of the integral will be independent of the path if we can find a function $f(x,y)$ that satisfies the gradient equation above.

You have

$\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}$

Integrate $\dfrac{\partial f}{\partial x}$ with respect to $x$ . You get

$\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx$
$f=x^2e^{-y}+g(y)$

Differentiate with respect to $y$ . You get

$\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]$
$2y-x^2e^{-y}=-x^2e^{-y}+g'(y)$
$2y=g'(y)$

Integrate both sides with respect to $y$ to arrive at

$\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy$
$y^2=g(y)+C$
$g(y)=y^2+C$

So you have

$f(x,y)=x^2e^{-y}+y^2+C$

The gradient is continuous for all $x,y$ , so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e$