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umka21 [38]
3 years ago
14

Write four numbers that could be ordered between these numbers 59 through 1,000

Mathematics
2 answers:
Blababa [14]3 years ago
7 0
100, 200, 300, and 400.
Mrrafil [7]3 years ago
3 0
99, 199, 299, 399,
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Solve the triangle A = 2 B = 9 C =8
VARVARA [1.3K]

Answer:

\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}

Then, given the sides a=2, b=9, and c=8.

\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}

For B:

\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}

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1 year ago
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Explanation:
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So to get the answer you do 100100100+303030-101010 which = 100302120 x 10 (the amount she/he gets for each subscriber) which is the amount of money Kiersten earned.

Hope this helped 

4 0
3 years ago
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