All categories

3 years ago

Only answer 28 ( I added this parentheses so that the thing would accept my question)

Mathematics

1 answer:

avanturin [10]3 years ago

3 0

Answer:

(a). Number of protons in a chlorine atom = 17

(b). Atomic number of sodium = 11

Step-by-step explanation:

Let the number of protons in a sodium atom = x

Then, the number of protons in a chlorine atom = x + 6

Now, it is given that the atomic number of an element is equal to the number of protons per atom.

So, atomic number of sodium (Na) = x

Atomic number of chlorine (Cl) = x + 6

It is given that the atomic number of chlorine is 5 less than twice the atomic number of sodium.

∴ atomic number of chlorine (Cl) = 2 × atomic number of sodium (Na) - 5

⇒ x + 6 = 2x - 5

⇒ x - 2x = -5 - 6

⇒ -x = -11

Cancelling the minus sign from both sides, we get

x = 11

(a). Number of protons in a chlorine atom = x + 6

= 11 + 6 (∵ x = 11)

= 17

(b). Atomic number of sodium = x = 11

You might be interested in

Help needed! Can I get some help with two questions?

Elodia [21]

Red

This is a circle with center (h, k) at (0,0) and radius (r) = 6

Formula: (x - h)² + (y - k)² = r²

Answer: x² + y² = 36

***************************************

Blue

This is a circle with center (h, k) at (2,2) and radius (r) = 1

Formula: (x - h)² + (y - k)² = r²

Answer: (x - 2)² + (y - 2)² = 1

***************************************

Green:

This is a circle with center (h, k) at (-2,-2) and radius (r) = 1

Formula: (x - h)² + (y - k)² = r²

Answer: (x + 2)² + (y + 2)² = 1

***************************************

Purple:

This is an ellipse with center (h, k) at (0,-2) and radius on the x-axis (a) = 3 and radius on the x-axis (b) = 1

Formula: $\frac{(x-h)^{2}}{a^{2}} +\frac{(y-k)^{2}}{b^{2}}=1$

Answer: $\frac{x^{2}}{9} +(y+2)^{2}=1$

***************************************

Black:

This is a parabola with vertex (h, k) at (0,-6) and vertical stretch (a) = -1

Formula: y = a(x - h)² + k

Answer: y = -x² - 6

7 0

2 years ago

Show that x+1 is a factor of f(x)=2x^4-x^3-35x^2+16x+48 . Then factor f(x) completely

Bond [772]

Answer:

callate

Step-by-step explanation:

4 0

3 years ago

How to round 831,756 nears hundred thousand

Anettt [7]

Answer:

800,000

Step-by-step explanation:

If the number was above 850,000 it would round up. If the number was 849,999 or below it would round down. In other words, you look at the number to the right of the hundred thousands place, or the number in the tens of thousands place, and if it is 5 or above you round up. 4 and below you round down.

4 0

2 years ago

Find the missing value in each figure below. What does “y” equal?

finlep [7]

Answer:

Step-by-step explanation:

The perpendicular is equal to 6. That's because the left triangle's missing angle is 180 - 45 -90 = 45

The angle in the right triangle is given as 52.

The cos(52) = adjacent side (which we just found to be 6) / y

Multiply both sides by y

y cos(52) = 6

cos(52) = 0.6157

Divide by sides by cos(52)

y = 6 / cos(52)

y = 6 / 0.6157

y = 9.76

8 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.)

5 0

3 years ago