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2 years ago

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Mathematics

1 answer:

Bad White [126]2 years ago

6 0

The net would be 51 feet long to stretch from one side of the court to the other

<h3>What are scaled drawings?</h3>

Scale drawings are drawings that shows a diagram with another proportional dimensions

The scale ratio of the diagram is given as:

Scale = 1 inch = 12 feet

The length of the net is given as:

Length = 4.25 inches

So, the actual length of the net is:

$Length = 4.25 \times 12$

Evaluate the product

$Length = 51$

Hence, the net would be 51 feet long, if stretched across the court

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A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\$

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$

Volume of the box at x= 1.59:

$V = \frac{1}{2} (5 – 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\$

Volume of the box at x= 0.53:

$V = \frac{1}{2} (5 – 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3$

The volume of the box is maximized when x = 0.53 cm

Therefore,

x = 0.53 cm

Maximum volume = 1.75 cm³

7 0

2 years ago

I need help please :’( Simplify -4x - x

timofeeve [1]

Answer:

-5x

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

For the following exercises, evaluate the expressions, given functions f, g, and h:

sertanlavr [38]

Answer:

- 16

Step-by-step explanation:

Evaluate f(2) by substituting x = 2 into f(x) , that is

f(2) = 3(2) - 3 = 6 - 3 = 3

Evaluate g(- 1) by substituting x = - 1 into g(x) , that is

g(- 1) = 8 - (- 1)² = 8 - 1 = 7

Thus

4f(2) - 4g(- 1)

= 4(3) - 4(7)

= 12 - 28

= - 16

7 0

2 years ago

Years of experience for each

garik1379 [7]

Around 45% since there is 5 with more than 11 years of experience and there is 11 total teachers 5/11 is around 45%

6 0

2 years ago

Read 2 more answers

100 POINT HELP PLEASE !!!!!!!!!!! NO LINKS

NikAS [45]

y = log(3x + 4) + 2

__________________________________________________________

Y-INTERCEPT: set x = 0

y = log(3(0) + 4) + 2

y = log(0 + 4) + 2

y = log( 4) + 2

Using calculator,

y = log( 4) + 2

y ≈ 2.60 when x = 0

__________________________________________________________

X-INTERCEPT: set y = 0

y = log(3x + 4) + 2

0 = log(3x + 4) + 2

-2 = log(3x + 4)

(it's simple. Always remember that the base is 10, even though there seems to be no "smaller number" next to log)

= 3x + 4

$10^{-2}$ - 4 = 3x

[ $10^{-2}$ - 4 ] /3 = x

x = -1.33 when y = 0

__________________________________________________________

<h3>ANSWER: y-intercept is 2.60 and x-intercept is -1.33</h3>

4 0

2 years ago