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3 years ago

Write a division problem that has a quotient of 1/20

Mathematics

1 answer:

alexdok [17]3 years ago

6 0

Answer:

5 divided by 100

Step-by-step explanation

5 divided by 100 is 0.05 but as a fraction, it's 1/20

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41x0x3 what property is this

Drupady [299]

The Multiplication Property of Zero.

It states "the product of any number and zero is zero". Even if there are more than one multiplier - it is still 0.

So 5 x 0 = 0

100 x 0 = 0

41 x 0 x 3 = 0

etc.

6 0

3 years ago

Read 2 more answers

the contractor demands a working temperature of 82 degrees Fahrenheit. what is the equivalent temperature in degrees Celsius

stiks02 [169]

Answer:

27.8 Celsius

Step-by-step explanation:

The equation to convert Fahrenheit and Celsius is:

F = (9/5)C + 32

82 = (9/5)C + 32 -- subtract 32 from both sides

50 = (9/5)C -- multiply both sides by 5

250 = 9C -- divide by 9

27.777 = C

round up to 27.8 C

7 0

3 years ago

Can someone please please PLEASEEE. Help me on this one

Furkat [3]

Answer:

256 total

Step-by-step explanation:

112 boys

144 girls

3 0

3 years ago

Find x in the triangles

Degger [83]

Tan 32=9/x
x=14.4

cos29=x/7
x=6.12

4 0

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

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

3 years ago