What is 1.5x minus 3y equals 6. Solve for Y

What is the volume of the cone

Umnica [9.8K]

Answer:

37.68

Step-by-step explanation:

use the formula pi(3.14)r^2 h/3

3 0

3 years ago

Which inequality models this problem?

lana [24]

Answer:

C

Step-by-step explanation:

$6500 is a one time purchase so there is no variable attached.

$550 and $900 per week is reoccurring so there will be a variable attached.

Since $900 is what she is making each week, this will be separate from her costs (can eliminate D).

To make a profit, her amount earned will need to be greater than her expenses, so the answer is C.

8 0

3 years ago

A store had 5 packs of paper for $7.80. How much would it cost if you were to buy 3 packs

valentina_108 [34]

To buy 3 packs, the cost would be $4.68

4 0

3 years ago

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

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

The limit is 0, so any choice of $x$ satisfies the convergence condition.

3 0

4 years ago

Segment BD is an altitude of triangle ABC. Find the area of the triangle. Triangle ABC with altitude BD is shown. Point A is at

pogonyaev

Where's the "triangle with alt. BD?" This problem can be solved without the diagram, but the solution would be easier with it.

BD is the altitude. Find the length of BD by finding the dist. between (-1,4) and (2,4); it is 2-(-1), or 3. |BD| = 3.

I've graphed the triangle myself and have found that the "base" of the triangle is the vertical line thru (2,1) and (2,6); its length is 6-1, or 5.

Thus, the area of this triangle is A = (b)(h) / 2, or A = (5)(3) / 2 = 10/3 square inches.

7 0

3 years ago

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