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3 years ago

Find the product and express it in scientific notation.

Mathematics

2 answers:

rodikova [14]3 years ago

8 0

The answer is b because you multiply 2.635 times 10 then subtract it by 6

Tresset [83]3 years ago

4 0

Answer:

Step-by-step explanation:

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PLEASE HELP

lukranit [14]

Answer:

Step-by-step explanation:

l = 34 cm

height = 30 cm

radius = 16 cm

surface area of cone = πr (l +r)

=3.14 *16 (34 + 16)

=50.24*50

=2512 cm^2

volume of a cone = $\frac{1}{3}$ πr^2h

= $\frac{1}{3}$ *3.14*16^2 *30

=1.0466*256*30

=8037.888

=8037.89 cm^3

3 0

3 years ago

HELP ASAP PLLLLLLLLLLS!

PolarNik [594]

Answer:

uummmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

Step-by-step explanation:

8 0

3 years ago

Determine the domain of the function f(x)=sqrt of x+3/(x+8)(x-2)

Karo-lina-s [1.5K]

Unclear. Did you mean

<span>f(x)= [sqrt of x+3]
--------------------
(x+8)

or did you mean
3
</span><span>f(x)= [sqrt of x] + ----------
x+8

I will focus on the first possibility. The domain of sqrt(x+3) is [-3,infinity).
The domain of x+3
------- is "x is not equal to -8."
x+8

Then the overall domain would be simply [-3, infinity).
</span>

7 0

3 years ago

Write the Taylor Series for f(x) = sin(x)center

tangare [24]

Answer:

Taylor series of sin(x) centered at x = 0.

Step-by-step explanation:

Taylor series expansion:

$\sum_{n=0}^N f^n(a)\displaystyle\frac{(x-a)^n}{n!}$

Here, f(x) = sin (x) and a = 0

$f(x) = \sin x, f(0) = 0\\f'(x) = \cos x, f'(0) = 1\\f''(x) = -\sin x, f''(0) = 0\\f'''(x) = -\cos x, f'''(0) = -1\\f^4(x) = \sin x, f^4(0) = 0\\f^5(x) = \cos x, f^5(0) = 0$

Putting all the values and expanding, we get,

$f(x) = f(a) + \displaystyle\frac{f'(a)(x-a)}{1!} + \displaystyle\frac{f''(a)(x-a)^2}{2!} + \displaystyle\frac{f'''(a)(x-a)^3}{3!} + \displaystyle\frac{f^4(a)(x-a)^4}{4!} + \displaystyle\frac{f^5(a)(x-a)^5}{5!} + ...\\\\= \sin 0 + \displaystyle\frac{x}{1!} + \displaystyle\frac{(0)x^2}{2!} + \displaystyle\frac{(-1)x^3}{3!} + \displaystyle\frac{(0)x^4}{4!} + \displaystyle\frac{(1)x^5}{5!} + ...$

Solving, we get

$\sin x = x - \displaystyle\frac{x^3}{3!} + \displaystyle\frac{x^5}{5!} - \displaystyle\frac{x^7}{7!} + ...\\\\\sin x = x - \displaystyle\frac{x^3}{6} + \displaystyle\frac{x^5}{120} - \displaystyle\frac{x^7}{5040} + ...$