Ask question

Ask question

All categories

alexandr402 [8]

3 years ago

10

Please help me.

1 answer:

sergey [27]3 years ago

6 0

I don’t know if you are in a

You might be interested in

How do i solve for the inequality, p=3.4?

Roman55 [17]

Answer:

Step-by-step explanation:

This is already solved, and this would not be considered a inequality since it has only a equal sign.

7 0

3 years ago

What is the answer.

Alenkasestr [34]

√11 √5 = √(11 * 5) = √55 = approx 7.42 (rounded)

4 0

4 years ago

Whats next 2 numbers in this pattern 4,20,100,500.....

Korolek [52]

The next two patterns are 2,500 12,500

7 0

4 years ago

Solve the system by elimination 1: x+5y-4z=-10, 2x-y+5z=-9, 2x-10y-5z=0 .. . 2. Solve the system by substitution: 2x-y+z=-4, z=5

Kryger [21]

1 )
1st and 2nd equation (multiply 1st by -2 ):
-2x - 10 y + 8 z = 20
2 x - y + 5 z = -9
---------------------------
- 11 y + 13 z = 11 / * ( -20 )
1st and 3rd:
- 2 x - 10 y + 8 z = 20
2 x - 10 y - 5 z = 0
-----------------------------
- 20 x + 3 z = 20 / * 11
----------------------------------
220 y - 260 z = -220
- 220 y + 33 z = 220
------------------------------
z = 0, y = -1, x = - 5
2 )
2 x - y + 5 = -4
- 2 x + 3 y - 5 = - 10
Substitution: y = 2 x + 9
- 2 x + 3(2 x + 9 ) = -5
4 x = - 32, x = -8, y = -7, z = 5
3 )
x + y + z = 11
1.5 x+3 y +1.5 z = 21
x = 2 y
------------------------
3 y + z = 11 / * (-2)
6 y + 1.5 z = 21
------------------------
- 6 y - 2 z = -22
6 y + 1.5 z = 21
------------------------
-0.5 z = -1, z = 2, y = 3, z = 6
Answer: the store used 6 pounds of peanuts, 3 pounds of almonds and 2 pounds of raisins.

8 0

4 years ago

If 30,000 cm2 of material is available to make a box with a square base and an open top, what is the largest possible volume (in

Cloud [144]

Answer:

The largest possible volume of the box is 2000000 cubic meters.

Step-by-step explanation:

The volume ( $V$ ), in cubic centimeters, and surface area ( $A_{s}$ ), in square centimeters, of the box with a square base are described below:

$A_{s} = l^{2}+h\cdot l$ (1)

$V = l^{2}\cdot h$ (2)

Where:

$l$ - Side length of the base, in centimeters.

$h$ - Height of the box, in centimeters.

By (2), we clear $h$ within the formula:

$h = \frac{V}{l^{2}}$

And we apply in (1) and simplify the resulting expression:

$A_{s} = l^{2}+ \frac{V}{l}$

$A_{s}\cdot l = l^{3}+V$

$V = A_{s}\cdot l -l^{3}$ (3)

Then, we find the first and second derivatives of this expression:

$V' = A_{s}-3\cdot l^{2}$ (4)

$V'' = -6\cdot l$ (5)

If $V' = 0$ and $A_{s} = 30000\,cm^{2}$ , then we find the critical value of the side length of the base is:

$30000-3\cdot l^{2} = 0$

$3\cdot l^{2} = 30000$

$l = 100\,cm$

Then, we evaluate this result in the expression of the second derivative:

$V'' = -600$

By Second Derivative Test, we conclude that critical value leads to an absolute maximum. The maximum possible volume of the box is:

$V = 30000\cdot l - l^{3}$

$V = 2000000\,cm^{3}$

The largest possible volume of the box is 2000000 cubic meters.

4 0

3 years ago