Math test! i need help!

What quantum numbers specify these subshells 5s 2p 3d?

Ivan

5s subshell represent that n is equal to 5 and l is equal to 0.
2p subshell represent that n is equal to 2 and l is equal to 1.
3d subshell represent that n is equal to 3 and l is equal to 2.
The numbers gives us the value of n and the letter gives us the value of l.
s means l = 0
p means l = 1
d means l = 2
f means l = 3

5 0

3 years ago

What is the volume of a rectangular prism with a length of 14 m, height of 6

anyanavicka [17]

Answer:

252m^3

Step-by-step explanation:

Multiply all the numbers.

5 0

2 years ago

Read 2 more answers

Would appreciate the help 24 points

andre [41]

Answer:

13) Rectangle

14) parallelogram

15) co ordinates areD (0,a) andE (c,b)

16) co ordinates are D(0,b) and E(a,0)

17)(-1,1) (4,1) (-1,4)

18)(0,0) (a,0) (a,a) (0,a)

Step-by-step explanation:

<h3>13)</h3><h3>Rectangle</h3>

The opposite sides of a rectangle are parallel and equal

left height = right height

(by using distance formula)

(0,3),(2,-2) = (5,5),(7,0)

√(2-0)²+(-2-3)² = √(7-5)²+(0-5)²

√(2)²+(-5)² = √(2)²+(-5)²

√4+25 = √(2)²+(-5)²

√29 = √29 hence equal

<h3>14)</h3><h3>parallelogram</h3>

The opposite sides of a rectangle are not equal

left height = right height

(by using distance formula)

(0,0),(3,-4) = (3,4),(7,0)

√(3-0)²+(-4-0)² = √(7-3)²+(0-4)²

√(3)²+(-4)² = √(5)²+(-4)²

√9+16 = √(25)+(16)

√(25) ≠√41

5 ≠ √41

<h3>15)</h3><h3>co ordinates are D(0,a) and E(c,b)</h3><h3>16)</h3><h3>co ordinates are D(0,b) and E(a,0) </h3>

since rhombus have equal opposite sides

6 0

3 years ago

What is the y value in this equation ?X(0)=(-1/3y)+6

snow_lady [41]

Answer:

18

Step-by-step explanation:

8 0

3 years ago

*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0

3 years ago