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3 years ago

Simply the expression -10 + 5y + 2x- 6y

Mathematics

1 answer:

algol [13]3 years ago

5 0

Answer: 2x−y−10

Step-by-step explanation:

Simplify the expression.

2x*1 = 2x

6y-5y = 1y = y

10 = 10

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Find the diameter of a cicle with an area of 240.48 π square millimeters?

seropon [69]

31.02 mm.

Step-by-step explanation:

Step 1:

The area of the given circle is 240.48 π sq mm

We need to find the diameter of the circle

Step 2:

The formula for obtaining the area of any circle is π*r² where r represents the radius of the circle

We know that the diameter of circle is 2 times its radius.

Hence equating the formula of the area of the circle to the given value we can find its radius. Then multiplying the radius by 2 , we get the diameter.

Step 3 :

Using the above method , we have

πr² = 240.48 π

=> r² = 240.48 π / π = 240.48

=> r = √240.48 = 15.51 approximately

Hence the diameter of the given circle is 2 * 15.51 = 31.02 mm.

6 0

3 years ago

3( 3v +5 ) = 39<br> Solve this problem

V125BC [204]

Answer:

V= 8/3

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

$\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

$-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)$

c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$