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Tems11 [23]
2 years ago
7

When I multiply 9x9 how would I go about counting by 9s?

Mathematics
1 answer:
11111nata11111 [884]2 years ago
5 0

Answer:

If you're looking for a shortcut, I usually find the number 1 less than what you are multiplying 9 by, this case 8, and then add a number that gets to 9.

So 9x6 would be:

5+4=9

54

Step-by-step explanation:

When multiplying by 9s the product's two numbers added together =9.

Sorry if this is confusing. Hope this helps though.

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\frac{b\: \times \:h}{2}

B refers to the base and H refers to the height of the triangle.

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Answer:

m + m = 36

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Step-by-step explanation:


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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

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