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lakkis [162]
3 years ago
15

HELPPP

Mathematics
2 answers:
densk [106]3 years ago
5 0

Answer:

Step-by-step explanation:

<u><em>Decompose each number:</em></u>

<em><u /></em>6) 90 = 9 \times 10\\\\7) 40 = 4 \times 10\\\\8)890 = 89 \times 10\\\\9) 300 = 3 \times 100\\\\10) 7000 = 7 \times 1000\\\\11)3700 = 37 \times 100\\\\

<em><u>Correct the error. Write the correct decomposition:</u></em>

<em><u /></em>12) 560 = 56 \times 10\\\\13) 4300 = 43 \times 100\\\\14) 6000 = 60 \times 100 \ \ or \ \ 6 \times 1000<em><u /></em>

Crazy boy [7]3 years ago
3 0

Answer:

Decompose each number:

\begin{gathered}6) 90 = 9 \times 10\\\\7) 40 = 4 \times 10\\\\8)890 = 89 \times 10\\\\9) 300 = 3 \times 100\\\\10) 7000 = 7 \times 1000\\\\11)3700 = 37 \times 100\\\\\end{gathered}

6)90=9×10

7)40=4×10

8)890=89×10

9)300=3×100

10)7000=7×1000

11)3700=37×100

Correct the error. Write the correct decomposition:

\begin{gathered}12) 560 = 56 \times 10\\\\13) 4300 = 43 \times 100\\\\14) 6000 = 60 \times 100 \ \ or \ \ 6 \times 1000\end{gathered}

12)560=56×10

13)4300=43×100

14)6000=60×100 or 6×1000

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Alex, Bob, and Claudia split 126 cm wire evenly among themselves. They then proceeded to cut their pieces of wire into smaller,
Westkost [7]
<h3>Answer:</h3>
  • Bob
  • 18 cm
<h3>Step-by-step explanation:</h3>

If Alex cut his wire 18 times, he ended up with 19 equal pieces. He kept 7, so has 7/19 of his 1/3 of the wire.

Bob cut his wire 20 times, so ended up with 21 pieces, of which he kept 9. So he has 9/21 = 3/7 of his 1/3 of the wire.

Claudia kept 1/13 of her 1/3 of the wire, so has the smallest piece.

Bob kept (3/7)·(1/3)·126 cm = 18 cm.

Alex kept (7/19)·(1/3)·126 cm ≈ 15.47 cm.

Bob kept the longest part of the original wire.

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2 years ago
The vertices of triangle RST are R (-7, 5), S(17, 5), T(5, 0). What is the perimeter of triangle RST?
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~\hfill ST=\sqrt{( -12)^2 + ( -5)^2}\implies \boxed{ST=13} \\\\\\ T(\stackrel{x_1}{5}~,~\stackrel{y_1}{0})\qquad R(\stackrel{x_2}{-7}~,~\stackrel{y_2}{5}) ~\hfill TR=\sqrt{(~~ -7- 5~~)^2 + (~~ 5- 0~~)^2} \\\\\\ ~\hfill TR=\sqrt{( -12)^2 + (5)^2}\implies \boxed{TR=13} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{\LARGE perimeter}}{24~~ + ~~13~~ + ~~13\implies \text{\LARGE 50}}~\hfill

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