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Virty [35]
4 years ago
6

The area of a circle is 28.26 square centimeters. What is its diameter (use 3.14 for ð)?

Mathematics
2 answers:
Inessa05 [86]4 years ago
8 0
The\ area\ of\ the\ circle:A_O=\pi r^2\\\\A_O=28.26\ cm^2;\ \pi\approx4.14\\\\subtitute\\\\3.14r^2=28.26\ \ \ |divide\ both\ sides\ by\ 3.14\\\\r^2=9\\\\r=\sqrt9\\\\r=3\ (cm)\\\\d-a\ diameter\\\\d=2r\\\\\boxed{d=2\cdot3cm=6cm}
choli [55]4 years ago
5 0
<span>Area of circle= pi*r^2
 28.26=pi*r^2
28.26/3.14=r^2
sqrt9=r
3=r
r=3
d=2r
2(3)
=6</span>
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PLEASE HELP
uranmaximum [27]

The coordinate for A, B, C, and D: (Current coordinates)

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Once we reflect acorss the y-axis the coordinates become:

A': (1, 4)

B': (5, 8)

C': (5, 4)

D': (4, 2)

In this situation the coordinates just become positive.

I hope this helps!

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4 years ago
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8 0
3 years ago
Read 2 more answers
Show work please<br> \sqrt(x+12)-\sqrt(2x+1)=1
Nesterboy [21]

Answer:

x=4

Step-by-step explanation:

Given \displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1, start by squaring both sides to work towards isolating x:

\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2

Recall (a-b)^2=a^2-2ab+b^2 and \sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}:

\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1

Isolate the radical:

\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}

Square both sides:

\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2

Expand using FOIL and (a+b)^2=a^2+2ab+b^2:

\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36

Move everything to one side to get a quadratic:

\displaystyle-\frac{1}{4}x^2+7x-24=0

Solving using the quadratic formula:

A quadratic in ax^2+bx+c has real solutions \displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}. In \displaystyle-\frac{1}{4}x^2+7x-24, assign values:

\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24

Solving yields:

\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}

Only x=4 works when plugged in the original equation. Therefore, x=24 is extraneous and the only solution is \boxed{x=4}

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Answer:

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