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san4es73 [151]
3 years ago
12

All parallelogram are rectangle?true or false​

Mathematics
2 answers:
Yuliya22 [10]3 years ago
7 0

Answer. False... All rectangles are parallelograms.

I hope this helps! I would really appreciate it if you would please mark me brainliest! Have a blessed day!

fredd [130]3 years ago
7 0

Answer: False

Step-by-step explanation:

Parallelogram does not include some of rectangle's property

  • Having four congruent angles
  • Having congruent diagonals

If you need a more graphical explanation, refer to the attachment below.

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On the first day of school each student is given $.50 to attend, on day two each student earns a dollar. On day three each stude
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Answer:

128 dollars

Step-by-step explanation:

4 0
4 years ago
[10 points] A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimens
MissTica

Answer:

  6 inches square by 3 inches high

Step-by-step explanation:

For a given surface area, the volume of an open-top box is maximized when it has the shape of half a cube. If the area were than of the whole cube, it would be 216 in² = 6×36 in².

That is, the bottom is 6 inches square, and the sides are 3 inches high.

_____

Let x and h represent the base edge length and box height, respectively. Then we have ...

  x² +4xh = 108 . . . . box surface area

Solving for height, we get ...

  h = (108 -x²)/(4x) = 27/x -x/4

The volume is the product of base area and height, so is ...

  V = x²h = x²(27/x -x/4) = 27x -x³/4

We want to maximize the volume, so we want to set its derivative to zero.

  dV/dx = 0 = 27 -(3/4)x²

  x² = (4/3)(27) = 36

  x = 6

  h = 108/x² = 3

The box is 6 inches square and 3 inches high.

_____

<em>Comment on maximum volume, minimum area</em>

In the general case of<em> an open-top box</em>, the volume is maximized when the cost of the bottom and the cost of each pair of opposite sides is the same. Here, the "cost" is simply the area, so the area of the bottom is 1/3 the total area, 36 in².

If the box has a <em>closed top</em>, then each pair of opposite sides will have the same cost for a maximum-volume box. If costs are uniform, the box is a cube.

7 0
3 years ago
What package of toothpaste has the lower unit price?
Maurinko [17]

Answer:

6.2 oz is lower

Step-by-step explanation:

6 0
2 years ago
Determine consecutive integer values of x between which each real zero is located.
frozen [14]

Answer:

1. x = -2 or x = sqrt(6) - 2 or x = -2 - sqrt(6)

2. x = -2.10947 or x = -0.484343 or x = 1.67884 or x = 2.91497

Step-by-step explanation:

Solve for x:

x^3 + 6 x^2 + 6 x - 4 = 0

The left hand side factors into a product with two terms:

(x + 2) (x^2 + 4 x - 2) = 0

Split into two equations:

x + 2 = 0 or x^2 + 4 x - 2 = 0

Subtract 2 from both sides:

x = -2 or x^2 + 4 x - 2 = 0

Add 2 to both sides:

x = -2 or x^2 + 4 x = 2

Add 4 to both sides:

x = -2 or x^2 + 4 x + 4 = 6

Write the left hand side as a square:

x = -2 or (x + 2)^2 = 6

Take the square root of both sides:

x = -2 or x + 2 = sqrt(6) or x + 2 = -sqrt(6)

Subtract 2 from both sides:

x = -2 or x = sqrt(6) - 2 or x + 2 = -sqrt(6)

Subtract 2 from both sides:

Answer: x = -2 or x = sqrt(6) - 2 or x = -2 - sqrt(6)

_________________________________________

Solve for x:

x^4 - 2 x^3 - 6 x^2 + 8 x + 5 = 0

Eliminate the cubic term by substituting y = x - 1/2:

5 + 8 (y + 1/2) - 6 (y + 1/2)^2 - 2 (y + 1/2)^3 + (y + 1/2)^4 = 0

Expand out terms of the left hand side:

y^4 - (15 y^2)/2 + y + 117/16 = 0

Subtract -3/2 sqrt(13) y^2 - (15 y^2)/2 + y from both sides:

y^4 + (3 sqrt(13) y^2)/2 + 117/16 = (3 sqrt(13) y^2)/2 + (15 y^2)/2 - y

y^4 + (3 sqrt(13) y^2)/2 + 117/16 = (y^2 + (3 sqrt(13))/4)^2:

(y^2 + (3 sqrt(13))/4)^2 = (3 sqrt(13) y^2)/2 + (15 y^2)/2 - y

Add 2 (y^2 + (3 sqrt(13))/4) λ + λ^2 to both sides:

(y^2 + (3 sqrt(13))/4)^2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = -y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2

(y^2 + (3 sqrt(13))/4)^2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = (y^2 + (3 sqrt(13))/4 + λ)^2:

(y^2 + (3 sqrt(13))/4 + λ)^2 = -y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2

-y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = (2 λ + 15/2 + (3 sqrt(13))/2) y^2 - y + (3 sqrt(13) λ)/2 + λ^2:

(y^2 + (3 sqrt(13))/4 + λ)^2 = y^2 (2 λ + 15/2 + (3 sqrt(13))/2) - y + (3 sqrt(13) λ)/2 + λ^2

Complete the square on the right hand side:

(y^2 + (3 sqrt(13))/4 + λ)^2 = (y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)))^2 + (4 (2 λ + 15/2 + (3 sqrt(13))/2) (λ^2 + (3 sqrt(13) λ)/2) - 1)/(4 (2 λ + 15/2 + (3 sqrt(13))/2))

To express the right hand side as a square, find a value of λ such that the last term is 0.

This means 4 (2 λ + 15/2 + (3 sqrt(13))/2) (λ^2 + (3 sqrt(13) λ)/2) - 1 = 8 λ^3 + 18 sqrt(13) λ^2 + 30 λ^2 + 45 sqrt(13) λ + 117 λ - 1 = 0.

Thus the root λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3)) allows the right hand side to be expressed as a square.

(This value will be substituted later):

(y^2 + (3 sqrt(13))/4 + λ)^2 = (y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)))^2

Take the square root of both sides:

y^2 + (3 sqrt(13))/4 + λ = y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)) or y^2 + (3 sqrt(13))/4 + λ = -y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) + 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2))

Solve using the quadratic formula:

y = 1/4 (sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) + sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 - 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) or y = 1/4 (sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) - sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 - 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) or y = 1/4 (sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 + 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13))) - sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13))) or y = 1/4 (-sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) - sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 + 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) where λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3))

Substitute λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3)) and approximate:

y = -2.60947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x - 1/2 = -2.60947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x - 1/2 = -0.984343 or y = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or x = -0.484343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x = -0.484343 or x - 1/2 = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or x = -0.484343 or x = 1.67884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x = -0.484343 or x = 1.67884 or x - 1/2 = 2.41497

Add 1/2 to both sides:

Answer: x = -2.10947 or x = -0.484343 or x = 1.67884 or x = 2.91497

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3 years ago
Martha is 25 years old, owns her car, but walks to work. Her collision
Shalnov [3]

Answer:B

Step-by-step explanation:

3 0
3 years ago
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