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Ne4ueva [31]
3 years ago
12

What is the least amount of wrapping paper needed to wrap a gift box that measures 8 inches by 8 inches by 10 inches. Explain

Mathematics
1 answer:
frozen [14]3 years ago
4 0

448 square inches.

The surface area of the box is 448 square inches so that is the least amount of paper needed to cover the box.

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If you used 1 drop of your sample and 4 drops of diluent what is your dilution ratio?
frez [133]

Answer:

1/5

Step-by-step explanation:

1 + 4 = 5

1/5

7 0
3 years ago
Read 2 more answers
The perimeter of a rectangle is 4040 yards. what are the dimensions of the rectangle with maximum area?
Ivanshal [37]
Let x be the length and y be the width2x + 2y = 40x + y = 20change that into y = mx + b formy= 20 – x
Area =xy = x (20-x) = 20x - x^2Area=-x^2+20xcomplete the square:Area=-(x^2 – 20x + 100) +100=-(x - 10)^2 + 100This is an calculation of a parabola that opens downward with vertex at (10,100), which means maximum area of 100 happens when x, the length=10)Dimensions of the rectangle with maximum area? 10 yds. by 10 yds., a square.
8 0
3 years ago
Help me out please T-T​
Anettt [7]

Answer:

The simplyfied version would be 19/4

Show of work:

(1/4)^-2 = 4^2

3 × 8^2/3 × 1 = 12

(9/16)^1/2 = 3/4

4^2 - 12 + 3/4

Convert elements to fractions:

-12 × 4 + 3

---------- ----

4 4

Since the denominators are equal combine the fractions:

-12 × 4 + 3

---------------

4

-12 × 4 + 3 = -45

= -45/4

=4^2 - 45/4

4^2 = 16

16 - 45/4

16 × 4 - 45. 16 × 4 - 45

--------- ----- ----------------

4 4. 4

-> 16 × 4 - 45 = 19

= 19/4

7 0
3 years ago
Read 2 more answers
4.0681 to the nearest thousand
Neporo4naja [7]
4.068 is the answer to it
5 0
3 years ago
Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt
brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

f'(x) = \dfrac{(-x + 2)^4}{dx}  = -4 \cdot (-x + 2)^3 = 0

∴ (-x + 2)³ = 0

x = 2

f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx}  = -12 \cdot (-x + 2)^2

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0
3 years ago
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