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3 years ago

100 kg of a fruit contained 90% water one week ago. How many kg of the fruit containing 80% water are there now?

Mathematics

1 answer:

VashaNatasha [74]3 years ago

4 0

90kg

It is just a simple proportional problem. If you subtract 10%, then you subtrat 10 kg.

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11. Describe and correct the error in evaluating the expression.<br> -2(-7)= -14

Ulleksa [173]

Answer:

Step-by-step explanation:

The left side

does not equal to the right side

−

, which means that the given statement is false.

False

7 0

3 years ago

What is the additive inverse of the complex number –8 + 3i?

Lady bird [3.3K]

Choices:
–8 – 3i
–8 + 3i
8 – 3i
<span>8 + 3i

The additive inverse of the complex number a + bi is -(a+bi) = -a - bi

In this case, a = -8 and b = 3i

a + bi = -8 + 3i
additive inverse is: -(-8+3i) = +8 - 3i

The additive inverse of the complex number -8 + 3i is 8 - 3i. The 3rd choice.

</span>

5 0

3 years ago

Read 2 more answers

Find the missing fraction. What is the ??? equal to? (the fractions aren't in order)

dimaraw [331]

Step-by-step explanation:

1/6 divided by 3=1/2.

I think this is the answer

5 0

2 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

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

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