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3 years ago

Put y’all life down below now

Mathematics

2 answers:

tatyana61 [14]3 years ago

7 0

Answer:

I don't have a life.....

Kaylis [27]3 years ago

7 0

Answer:

I am an Undertale trash. so this is my life.

Step-by-step explanation:

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A boat can travel 110 kilometers on 110 liters of gasoline. How far can it travel on 59 liters?

topjm [15]

59 kilometers because 110/110 is equal to 59/59

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3 years ago

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The area of the rectangle below is 18 square units. What is the width of the rectangle? Write and solve a division equation to f

kherson [118]

Answer:

24/5 or 4 and 4/5

Step-by-step explanation:

3 and 3/4=15/4

18 divided by 15/4=4 and 4/5

4 and 4/5=24/5

24/5 times 15/4=18

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3 years ago

This angle cuts out 1/4 of the circle. Find the measure of the angle.

kykrilka [37]

Answer:

90 degrees

Step-by-step explanation:

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3 years ago

Describe the geometrical transformation and write the approximate value with working

lianna [129]

Answer:

$\boxed{\text{a) Translation 5 units up; b)}\approx 110}\\$

Step-by-step explanation:

a) Mapping the graphs

Mapping one graph (A) onto another (B) means that each vertical line through a point in A intersects the graph of the B at only one point.

In the diagram, the graph of ƒ(x) = √(x² + 9) is the blue line and the graph of g(x) = 5 + √(x² + 9) is in green.

Each vertical line through ƒ(x) intersects g(x) in only one point.

For example, a vertical line from (4, 0) intersects ƒ(x) at (4, 5) and g(x)

at (4, 10).

Every point in g(x) is five units higher than the corresponding point in ƒ(x).

Thus, mapping ƒ(x) onto g(x) is a translation of five units upward.

b) Integration

$\int_{2}^{11}{(\sqrt{x^{2}+9}})dx\approx 65\\$

$\int_{2}^{11}{(5 + \sqrt{x^{2}+9})}dx \approx [5x]_2^{11}+65\\$

$\approx (55 - 10) + 65 \approx 45 + 65\\$

$\boxed{\approx 110}\\$

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3 years ago

3) Postcards: Marcie wants to make a box to hold her postcard collection from a piece of

klasskru [66]

The volume of the box is the amount of space in it.

The dimension of the postcard is:

$\mathbf{Length = 10}$

$\mathbf{Width = 18}$

Let x represents the length of the cut.

So, the dimension of the box is:

$\mathbf{Length = 10 -2x}$

$\mathbf{Width = 18 -2x}$

$\mathbf{Height = x}$

(a) The function that represents the volume of the box

This is calculated as:

$\mathbf{V(x) = Length \times Width \times Height}$

So, we have:

$\mathbf{V(x) = (10 - 2x) \times (18 - 2x) \times x}$

Expand

$\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

Hence, the volume function is: $\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

(b) The dimension that maximizes the volume

We have:

$\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

Differentiate

$\mathbf{V'(x) = 180- 112x + 12x^2}$

Set to 0

$\mathbf{180- 112x + 12x^2 = 0}$

Using a calculator, we have:

$\mathbf{x = 7.3,2.1}$

The value x = 7.3 is greater than the dimension of the box.

So, we have:

$\mathbf{x = 2.1}$

Recall that:

$\mathbf{Length = 10 -2x}$

$\mathbf{Width = 18 -2x}$

$\mathbf{Height = x}$

So, we have:

$\mathbf{Length = 10 -2 \times 2.1 = 5.8}$

$\mathbf{Width = 18 -2 \times 2.1 = 13.8}$

$\mathbf{Height = 2.1}$

Hence, the dimension that maximizes the volume is 5.8 by 13.8 by 2.1

(c) The maximum volume

This is calculated as the product of the dimension of the box

So, we have:

$\mathbf{Volume = 5.8 \times 13.8 \times 2.1}$

$\mathbf{Volume = 168.1}$

Hence, the maximum volume of the box is 168.1 cubic inches