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Snezhnost [94]
3 years ago
5

6. The rungs of a ladder are each perpendicular to one side of the ladder. What can you conclude about the rungs of the ladder?

Justify your answer by stating the theorem you used. ​
Mathematics
1 answer:
zubka84 [21]3 years ago
5 0

Answer: You can conclude that the ladder rungs are parallel using the Parallel Transversal Theorem.

Step-by-step explanation: Picture a ladder, all of the rungs are horizontal, but the sides are vertical. So all of the rungs are perpendicular to the sides, making right angles. If all of the rungs make 90 degree angles to the sides, that means that all of the rungs are straight with no angle. This makes sure that the rungs will not hit each other, so they are parallel.

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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
I WILL MARK YOU BRAINLYEST if you answer all of my problems !!!!!!!!!!!
Romashka [77]

First, let's convert each line to slope-intercept form to better see the slopes.

Isolate the y variable for each equation.

2x + 6y = -12

Subtract 2x from both sides.

6y = -12 - 2x

Divide both sides by 6.

y = -2 - 1/3x

Rearrange.

y = -1/3x - 2


Line b:

2y = 3x - 10

Divide both sides by 2.

y = 1.5x - 5


Line c:

3x - 2y = -4

Add 2y to both sides.

3x = -4 + 2y

Add 4 to both sides.

2y = 3x + 4

Divide both sides by 2.

y = 1.5x + 2


Now, let's compare our new equations:

Line a: y = -1/3x - 2

Line b: y = 1.5x - 5

Line c: y = 1.5x + 2

Now, the rule for parallel and perpendicular lines is as follows:

For two lines to be parallel, they must have equal slopes.

For two lines to be perpendicular, one must have the negative reciprocal of the other.

In this case, line b and c are parallel, and they have the same slope, but different y-intercepts.

However, none of the lines are perpendicular, as -1/3x is not the negative reciprocal of 1.5x, or 3/2x.

<h3><u>B and C are parallel, no perpendicular lines.</u></h3>
8 0
3 years ago
What is the equation of the line that has a slope of -3 and passes through the point (4,6)?
garri49 [273]

Answer:

y = -3x + 18

Step-by-step explanation:

y = mx + b

m = -3

6 = (-3) * 4 +b

b = 6 - (-3) * 4

b = 6 +12

b = 18

y = -3x + 18

6 0
2 years ago
What is the domain of y = √x^2-1?
andriy [413]
X can be -1 or 1 for the square root to be 0 

The correct option is a
7 0
3 years ago
Read 2 more answers
What's greater -5/7 or -2/7
nadezda [96]

I think -2/7.

-2/7 is closer to 0 which is neither positive nor negative.

-5/7 is closer to a whole negative number.

4 0
2 years ago
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