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3 years ago

Question 8 options:

Mathematics

1 answer:

shusha [124]3 years ago

4 0

Answer:

volume of cone=1/3 πr²h=1/3 ×π×5²×15=125πcm³

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Kite QRST has a short diagonal of QS and a long diagonal of RT. The diagonals intersect at point P. Side QR= 5 m and diagonal QS

svet-max [94.6K]

Answer:

Given the statement: Kite QRST has a short diagonal of QS and a long diagonal of RT. The diagonals intersect at point P.

Properties of Kite:

The diagonals are perpendicular

Two disjoint pairs of consecutive sides are congruent by definition of kite

One diagonal is the perpendicular bisector to the other diagonal.

It is given that: Side QR = 5m and diagonal QS = 6m.

Then, by properties of kite:

$QP = \frac{1}{2}QS$

Substitute the value of QS we get QP;

$QP = \frac{1}{2}(6)$ = 3 m

Now, in right angle $\triangle RPQ$

Using Pythagoras theorem:

$QR^2= RP^2 +QP^2$

Substitute the given values we get;

$(5)^2= RP^2 +(3)^2$

$25= RP^2 +9$

Subtract 9 from both sides we get;

$16= RP^2$

Simplify:

$RP = \sqrt{16} = 4 m$

Therefore, the length of segment RP is, 4m

5 0

3 years ago

Read 2 more answers

Solve the equation. |3 − w| = −2

gizmo_the_mogwai [7]

Absolute value represents distance. Distances can't be negative.

So it's impossible to have |3-w| be some negative value.

Which is why $|3-w| = -2$ has no solutions

3 0

3 years ago

Read 2 more answers

Please help me with this

kondaur [170]

For number 1a. the answer is 7 to 26 or
7:26 or
$\frac{7}{26}$
For number 2a. the answer is 8 to 28 or
8:28 or
$\frac{8}{28} or \frac{2}{7}$
For number 3a. the answer is 12 to 13 or
12:13 or
$\frac{12}{13}$
For number 4a. the answer is 19 to 12 or
19:12 or
$\frac{19}{12}$
For number 5a. the answer is 29 to 10 or
29:10 or
$\frac{29}{10}$
For number 6a. the answer is 11 to 28 or
11:28 or
$\frac{11}{28}$

I just did your homework

4 0

3 years ago

Find the domain and range of the function graphed in the picture.

algol [13]

The domain is the set of allowed x inputs, or x coordinates of a function. In this case, any point on the curve has an x coordinate that is 4 or smaller.

Therefore, the domain is the set of numbers x such that $x \le 4$

To write this in interval notation, we would write $(-\infty, 4]$ This interval starts at negative infinity and stops at 4. We exclude infinity with the curved parenthesis and include 4 with the square bracket.

-----------------------------------------------------------------------

The range is the set of possible y outputs. Every point on this curve has a y coordinate that is either 0 or it is larger than 0.

The range is the set of y values such that $y \ge 0$

In interval notation, it would be written as $[0, \infty)$ This time we start at 0 (including this endpoint) and "stop" at infinity

note: we always use curved parenthesis at positive or negative infinity because we cannot reach either infinity

7 0

3 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