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Galina-37 [17]
3 years ago
8

Which estimate is the closest to actual value of (2.99548) (1.8342)?

Mathematics
2 answers:
anzhelika [568]3 years ago
6 0

Answer:

B 5.5 I had it on a exit ticket and I got it right

Step-by-step explanation:

Jet001 [13]3 years ago
4 0
The answer will be B ~5.5
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Is 11/16 bigger than 3/4
Goryan [66]
No 3/4 is bigger

if you multiply the top and bottom of 3/4 by 4, you get 12/16.

11/16 <12/16
8 0
3 years ago
Simplify this problem!!!
Nonamiya [84]

Answer:

C

Step-by-step explanation:

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4 0
2 years ago
Read 2 more answers
Pls help me i need help
Llana [10]

Answer:

3x was subtracted from the left side, but 3x was subtracted from the right side. The Subtract Property of Equality states that you can subtract the same number from each side and the equation will remain true. But 3x and 3 are not the same number (unless x is 1).

x = -8/5

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

  • (7x+3x)+(21)=5
  • 10x + 21 = 5

Step 2: Subtract 12 from both sides.

  • 10x + 21 - 21 = 5 - 21
  • 10x = -16

Step 3: Divide both sides by 10.

  • \frac{10x}{10} = \frac{-16}{10}
  • x = -\frac{8}{5}
8 0
3 years ago
Read 2 more answers
A+B=37<br> A-B=9<br> Dau coroana
azamat
12-3=9
30+7=37
I believe you just plug in any two numbers that will come out to 9 when subtracted or 37 when added.
4 0
3 years ago
Find the area of the shaded region ​
o-na [289]

so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.

\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)

A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill

\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}

5 0
2 years ago
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