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xz_007 [3.2K]
2 years ago
7

The perimeter of a parallelogram must be no less than 40 feet. The length of the rectangle is 6 feet. What are the possible meas

urements of the width? Write an inequality to represent this problem. Use w to represent the width of the parallelogram. [Hint: The formula for finding the perimeter of a parallelogram is P = 2 l + 2 w . What is the smallest possible measurement of the width? Justify your answer by showing all your work.
Mathematics
1 answer:
Alekssandra [29.7K]2 years ago
8 0

Answer:

Step-by-step explanation:

so what you nedd otj fkhy is and the anwser is 799

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Consider the first five steps of the derivation of the Quadratic Formula.
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Answer:

Full proof below

Step-by-step explanation:

\displaystyle ax^2+bx+c=0\\\\ax^2+bx=-c\\\\x^2+\biggr(\frac{b}{a}\biggr)x=-\frac{c}{a}\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\biggr(\frac{b}{2a}\biggr)^2=-\frac{c}{a}+\biggr(\frac{b}{2a}\biggr)^2\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=\frac{b^2-4ac}{4a^2}\\ \\x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

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2 years ago
<img src="https://tex.z-dn.net/?f=4%20%2B%20%20%7B5%7D%5E%7B2%7D%20" id="TexFormula1" title="4 + {5}^{2} " alt="4 + {5}^{2} "
Tju [1.3M]
5 squared is 25. 25+4. Answer is 29
5 0
4 years ago
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Solve and explination
Oksana_A [137]

Answer:

160ft

Step-by-step explanation:

To find the area of something, you need to do base x height. Which is,

8 x 20 = 160

Please give me brainlist!

6 0
3 years ago
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How do I find the total perimeter of a odd shape
Stella [2.4K]
The area of the irregular shape is the area of the two rectangles added together. Use the measure calculate feature of Geometer's Sketchpad to add the areas. The perimeter of the irregular shape is equal to the sum of the six segment around the outside of the figure.
3 0
3 years ago
Abby is buying a widescreen TV that she will hang on the wall between two windows. The windows are 36 inches apart, and wide scr
Musya8 [376]

Answer:

D < 40.2 inches

Step-by-step explanation:

The maximum width of the TV must be 36 inches. Since TVs are approximately twice as wide as they are tall, the maximum height is 18 inches.

The diagonal of a TV can be determined as a function of its width (w) and height (h) as follows:

d^2=h^2+w^2\\d=\sqrt{18^2+36^2}\\d= 40.2\ in

Therefore, the diagonal must be at most 40.2 inches.

Since the answer choices were not provided with the question, you should choose the biggest value that is under 40.2 inches.

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