How do I solve 8 - 4x + 5y when x = -2 and y = 6

Using the drawing tool, graph the two inequalities on the number line. Take a screenshot of each number line and submit it in th

masya89 [10]

Zero on number lines is neither negative nor positive, it is used as a sort of divider number;)

Step-by-step explanation: this picture should help explain a little

4 0

2 years ago

Read 2 more answers

Solve each pair of simultaneous equations by substitution method

WARRIOR [948]

Answer:

x = 1 y = -4

Step-by-step explanation:

- Plug the value of y into the other equation.

x + 3y = -11

x + 3(-4x) = -11

x - 12x = -11

-11x = -11

x = 1

- Now substitute the value of x into any equation.

y = -4x

y = -4(1)

y = -4

3 0

2 years ago

Read 2 more answers

What number is 36% of 85

Wittaler [7]

I think it would be 30.6

4 0

3 years ago

Read 2 more answers

MARSVECTORCALC6 3.4.020. My Notes A rectangular box with no top is to have a surface area of 64 m2. Find the dimensions (in m) t

ioda

Answer:

We would have

$l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}$

where " l " is length, " w" is width and "h" is height.

Step-by-step explanation:

Step 1

Remember that

Surface area for a box with no top = $lw+2lh+2wh = 64$

where " l " is length, " w" is width and "h" is height.

Step 2.

Remember as well that

Volume of the box = $l*w*h$

Step 3

We can now use lagrange multipliers. Lets say,

$F(l,w,h) = lwh$

and

$g(l,w,h) = lw+2lh+2wh = 64$

By the lagrange multipliers method we know that

$\nabla F = \lambda \nabla g$

Step 4

Remember that

$\nabla F = (wh,lh,lw)$

and

$\nabla g = (w+2h,l+2h , 2w+2l)$

So basically you will have the system of equations

$wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)$

Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get

$lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)$

Then you would get

$l\lambda (w+2h) = w\lambda (l+2h) = h\lambda (2w+2l)$

You can get rid of $\lambda$ from these equations and you would get

$lw+2lh = lw+2wh = 2wh+2lh$

And from those equations you would get

$l = w =2h$

Now remember the original equation

$lw+2lh+2wh = 64$

If we plug in what we just got, we would have

$l^{2} + l^{2} + l^2 = 64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}$

7 0

3 years ago

Consider a melody to be 7 notes from a single piano octave, where 2 of the notes are white key notes and 5 are black key notes.

igor_vitrenko [27]

Answer:

35,829,630 melodies

Step-by-step explanation:

There are 12 half-steps in an octave and therefore $12^7$ arrangements of 7 notes if there were no stipulations.

Using complimentary counting, subtract the inadmissible arrangements from $12^7$ to get the number of admissible arrangements.

$\displaystyle \_\_ \:B_1\_\_ \:B_2\_\_ \:B_3\_\_ \:B_4\_\_ \:B_5\_\_$

$B_1$ can be any note, giving us 12 options. Whatever note we choose, $B_2, B_{...}$ must match it, yielding $12\cdot 1\cdot 1\cdot 1\cdot 1=12$ . For the remaining two white key notes, $W_1$ and $W_2$ , we have 11 options for each (they can be anything but the note we chose for the black keys).

There are three possible arrangements of white key groups and black key groups that are inadmissible:

$WWBBBBB\\WBBBBBW\\BBBBBWW$

White key notes can be different, so a distinct arrangement of them will be considered a distinct melody. With 11 notes to choose from per white key, the number of ways to inadmissibly arrange the white keys is $\displaystyle\frac{11\cdot 11}{2!}$ .

Therefore, the number of admissible arrangements is:

$\displaystyle 12^7-3\left(\frac{12\cdot 11\cdot 11}{2!}\right)=\boxed{35,829,630}$

6 0

2 years ago