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3 years ago

Solve -4.8=d-3.9Enter only the value of the variable

Mathematics

2 answers:

scoundrel [369]3 years ago

6 0

Answer:

D= -0.9

Step-by-step explanation:

Mrac [35]3 years ago

5 0

Answer:

Step-by-step explanation:

-4.8 = d - 3.9

+3.9 +3.9

-0.9 = d

Check:

-4.8 = -0.9 -3.9

-4.8 = -4.8

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The graph of a system of parallel lines will have no solutions.

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This is true. Parallel lines have the same slope, and therefore will not cross unless they are the same line. A 'solution' is the point where two or more lines cross through each other. When there is a system of parallel lines, there will be no solutions. If it is the same line, there will be an infinite number of solutions, technically.

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3 years ago

Solve the simultaneous equations: 5x + 3y = 9 7x - 2y = 25 x=__ y=__

kondor19780726 [428]

Answer:

x= 3, y= -2

Step-by-step explanation:

5x+3y= 9

x= 25÷7 + 2÷7 y

5(25÷7 +2÷7 y)

5( 25÷7 +2÷7 y ) + 3y= 9

y= -2

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5 0

3 years ago

Find the slope of the line containing the two points. (2,5),(-3,-2) Please show work

HACTEHA [7]

Answer:

7/5

Step-by-step explanation:

To solve this problem we will use the formula to find the slope of a line given two points (x₁, y₁), (x₂, y₂)

The formula is:

$m= \frac{y{_2} -y_{1}}{x_{2}-x_{1} } \\\\$

So, we're going to use the points (2,5), (-3, -2) in this formula.

$m= \frac{-2-(5)}{-3-(2)} \\m=\frac{-2-5}{-3-2}\\m=\frac{-7}{-5} \\m=\frac{7}{5}$

Therefore, the slope is 7/5

8 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