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4 years ago

Which of the following correctly

Mathematics

1 answer:

ikadub [295]4 years ago

3 0

Answer: 0.075, 0.75%, 3/4

Step-by-step explanation:

3/4 and 0.75% are the same value but expressed in different ways. 0.0075 is less than them.

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If the length of each side of a cuboid decreases by 20%, find the percentage decrease in its volume.

just olya [345]

Answer:

Step-by-step explanation:

(1+25 /100) (1-20/100) (1-50/100) <1

5/4 x 4/5 x 1/2 <1

Decrease in volume (in percent)

(1+25 /100) (1-20/100) (1-50/100) x 100

=48.8%

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3 years ago

Guillermo saved 25% of his paycheck last week. His paycheck was $1000. How much money did he save?

Bingel [31]

He saved 250 dollars.

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3 years ago

Read 2 more answers

Mitch is replacing all the light bulbs in his home with energy efficient LED light bulbs. How many packages of 4 will he need to

lapo4ka [179]

C) 22

86/4 = 21.5
*5 Or more, raise the score*
(rounds to 22)
you should have 2 lightbulbs left over but you would be 2 lightbulbs short if you were to get 21 packages as opposed to 22

6 0

4 years ago

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

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

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