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2 years ago

F(x,y,z)=66+5y+7+5t-9

1 answer:

5 0

1) X + Y = 10
2) X + Z = 20
3) Y + Z = 24
what is X + Y + Z = ?

From equation 1, x = 10-y
Plug 10-y in place of x in equation 2

10-y+z = 20 subtract 10 from both sides
-y + z = 10

We now have 2 equations with just y and z

-y + z = 10 new equation
y + z = 24 add the 2 equations
--------------
2z = 34
z = 34/2
z = 17

We now have the value of z.
Plug that into equation 3

y + z = 24
y + 17 = 24
y = 24-17
y = 7

We now have the value of both z and y.
Plug the value for y into equation 1 to find x

x + y = 10
x + 7 = 10
x = 10-7
x = 3

Your values are x = 3, y = 7, z = 17

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The diagram shows a solid formed by joining a hemisphere, of radius rcm, to a cylinder, of radius rcm and height hcm. The total

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Answer:

Step-by-step explanation:

The total surface area of the sphere = 2πr²

Total surface area of the cylinder = 2πr²+2πrh

Total surface area of the solid = 4πr²+2πrh

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3 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

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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

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

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