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3 years ago

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Which number comes first if the numbers 2.001, 0.0035, 0.0005, 4.56094 are arranged in order from most to fewest significant dig

its?
A. 0.0005
B. 0.0035
C. 2.001
D. 4.56094

2 answers:

deff fn [24]3 years ago

4 0

Answer:

D.

Step-by-step explanation:

PLZ give me brainliest

grigory [225]3 years ago

4 0

D. 4.56094 I hope that helps you good luck

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Find the measures of the numbered angles.

Arte-miy333 [17]

1 = 90
2 = 65
3 = 65
4 = 25

90+25=115
180-115=65

have a merry christmas :)

3 0

2 years ago

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

3 years ago

Help asap my last question GIVING BRAINLIST!!!!

Alex

Answer:

1) $x < 0$

2) $x > 0$

3) $x < 10$

4) $x < -\frac{31}{55}$

5) $x \geq 0$

6) $x \geq -1$

Step-by-step explanation:

1) $-3 > 2x -3$

$2x- 3 < -3$

$2x - 3 + 3 < -3 +3$

$2x < 0$

$\frac{2x}{2} < \frac{0}{2}$

$x < 0$

2) $5 > -x + 5$

$-x + 5 < 5$

$- x + 5 -5 < 5 - 5$

$-x < 0$

$\frac{-x}{-1} < \frac{0}{-1}$

$x > 0$

3) $\frac{x}{5} -2 < 0$

$5 \times (\frac{x}{5} - 2 ) \times 5 < 0$

$x - 10 < 0$

$x -10 + 10 < 0 +10$

$x < 10$

4) $-5x + 11 > 31$

$-5x \cdot 11 < 31$

$-55x < 31$

$\frac{-55x}{-55} > \frac{31}{-55}$

$x < -\frac{31}{55}$

5) $\frac{1}{2}x - 4 \geq -4$

$\frac{1}{2}x -4 +4 \geq -4 +4$

$\frac{1}{2}x \geq 0$

$\frac{\frac{1}{2}x }{\frac{1}{2} } \geq \frac{0}{\frac{1}{2} }$

$x \geq 0$

6) $-4x + 2 \leq 6$

$-4x +2 - 2 \leq 6 -2$

$-4x < 4$

$\frac{-4x}{-4} \leq \frac{4}{-4}$

$x \geq -1$

6 0

3 years ago

Find p(4) when p=.80

deff fn [24]

Answer:

3.2

Step-by-step explanation:

0.8(4) = 3.2

3 0

3 years ago

Replace * with a monomial so that the expression represents the square of a binomial: *+24x+9

Shkiper50 [21]

Answer:

* is $7x^{2}$

Step-by-step explanation:

(7x + 3)(x + 3)

= $7x^{2}$ + 24x + 9

5 0

3 years ago

Read 2 more answers