All categories

3 years ago

Help! Giving brainliest.

Mathematics

2 answers:

VMariaS [17]3 years ago

6 0

<u>Answer:</u>

a) Steven is incorrect.

b) Steven is incorrect, because he claims that, in other words, 15 units squared = 18 units squared. 18 > 15, so the statement that 15 = 18 is false. 15 does not equal 18.

<u>Step-by-step explanation:</u> Steven claims that the area of rectangle A is double the area of square B. The area of rectangle A is 15 units squared, because you get a product of 15 when you multiply the side lengths 5 and 3. The area of square B is 9 units squared, because you get a product of 9 when you multiply the side lengths 3 and 3. In other words, Steven is claiming that the area of rectangle A is not 15 units squared, but 18 units squared (this is double the area of square B, as 18 is the product of 9 and 2), and that statement is false, because we know that 18 > 15. Therefore, 18 cannot equal 15; the figure that is made from the value of double the area of square B (18 units squared) is larger than rectangle A (15 units squared) - this reinforces the fact that 18 > 15, and hence cannot be equal to 15.

Daniel [21]3 years ago

4 0

Answer:

The area of rectangle A is length x breadth

= 3x5

= 15 units^2

The area of square B is side x side

= 3x3

= 9 units^2

Therefore,

The area of the rectangle is not double of the area of the square because 9x2 is not equal to 15. Steven isn't correct.

You might be interested in

If g(x)= 4, then g(5x) = ?

Neporo4naja [7]

Answer:

g(5x)= 20

Step-by-step explanation:

g(x)=4 is basically saying that 1 x is equal to 4.

g(5x) wants to know what 5 times 1 x is equal to (5 times 4 (4 is equal to 1 x)

g(5x)= 5(x=4)

g(5x)= 20

4 0

3 years ago

For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:

Oksi-84 [34.3K]

The half-range sine series is the expansion for $f(t)$ with the assumption that $f(t)$ is considered to be an odd function over its full range, $-1$ . So for (a), you're essentially finding the full range expansion of the function

$f(t)=\begin{cases}2-t&\text{for }0\le t$

with period 2 so that $f(t)=f(t+2n)$ for $|t|$ and integers $n$ .

Now, since $f(t)$ is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

$f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L$

where

$b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt$

In this case, $L=1$ , so

$b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt$
$b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{4-2(-1)^n}{n\pi}$

The half-range sine series expansion for $f(t)$ is then

$f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t$

which can be further simplified by considering the even/odd cases of $n$ , but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming $f(t)$ is even/symmetric across its full range. In other words, you are finding the full range series expansion for

$f(t)=\begin{cases}2-t&\text{for }0\le t$

Now the sine series expansion vanishes, leaving you with

$f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L$

where

$a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt$

for $n\ge0$ . Again, $L=1$ . You should find that

$a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3$

$a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt$
$a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}$
$a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}$

Here, splitting into even/odd cases actually reduces this further. Notice that when $n$ is even, the expression above simplifies to

$a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0$

while for odd $n$ , you have

$a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}$

So the half-range cosine series expansion would be

$f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t$

Attached are plots of the first few terms of each series overlaid onto plots of $f(t)$ . In the half-range sine series (right), I use $n=10$ terms, and in the half-range cosine series (left), I use $k=2$ or $n=2(2)-1=3$ terms. (It's a bit more difficult to distinguish $f(t)$ from the latter because the cosine series converges so much faster.)