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zheka24 [161]
3 years ago
9

A=14 and b=6

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
3 0

Answer:

#1. Insert the given values into the equation:

Simplify multiplication from left to right:

Simplify subtraction:

#2. Insert the given values into the equation:

Use PEMDAS. Simplify multiplication:

Simplify division:

#3. Insert given values into the equation:

Use PEMDAS. Simplify parentheses:

Use PEMDAS. Simplify multiplication:

Simplify subtraction:

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For each ordered pair (x,y) , determine whether it is a solution to the inequality
Leviafan [203]
The answer are below, the work is show step by step.

8 0
3 years ago
Please help asap! (:
MAXImum [283]
Hello!

You can only add variables with the same base and exponent

8u^3 + 4u^3 = 12u^3
8u^2 + 0 = 8u^2
0 + -6u = -6u
6 + 3 = 9

Put the sums together

12u^3 + 8u^2 - 6u + 9

The answer is D) 12u^3 + 8u^2 - 6u + 9

Hope this helps!

6 0
3 years ago
Read 2 more answers
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
Last night, Christina received 4 messages from Kyle, 11 from Christina, 5 from Steven, 9 from Kevin, and 24 from Natalie. What i
bulgar [2K]
This = P(message is from Steven) =  number of messages from steven / total number of messages  from previous night

=  5 / (4+11+5+9+24) =   5/53 Answer
7 0
3 years ago
Easy math please help i'll mark brainliest
arlik [135]

i believe these are the answers:

1. b

2. a

4 0
3 years ago
Read 2 more answers
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