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nadya68 [22]
3 years ago
7

Which of the following sets of numbers could be the sides of a triangle

Mathematics
2 answers:
eimsori [14]3 years ago
5 0
Longest side must be smaller than the sum of other two smaller sides, so
only 2nd option fits best

In short, Your Answer would be Option B

Hope this helps!
anyanavicka [17]3 years ago
5 0
So bascically what you would do is you would do A^2+B^2=C^2 for every single one of them to find which one best fits the equation 6 squared +8 squared =15 squared 36+64=225 so that would be 100 which does not equal 225 therefore A is not the answer. 9 squared+ 13 squared= 17 squared 81+169= 289  250 does not equal 289 so therefore B is not the answer. 2 squared + 3 squared= 6 squared. 4+9=36. 13 does not equal 36 so therefore C is not the answer but because there are no more answers D is left by process of elimination
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3 years ago
Choose the option with the proper number of sig figs
marishachu [46]
\text{ 2 }\times10^8

Explanation:\begin{gathered} \text{Given:} \\ \frac{9\text{ }\times10^9}{4.5\text{ }\times10^1} \end{gathered}

let's break down the expression to get the final result:

\begin{gathered} \frac{9\text{ }\times10^9}{4.5\text{ }\times10^1}\text{ = }\frac{9}{4.5}\times\text{ }\frac{10^9}{10^1} \\ \frac{9}{4.5}\text{ = }\frac{9\text{ }\times\text{ 10}}{4.5\text{ }\times\text{ 10}}\text{ = }\frac{90}{45} \\ \frac{9}{4.5}\text{ = }2 \\  \\ \frac{10^9}{10^1}\colon\text{ when we divide exponents with same base, } \\ we\text{ take one of the base and combine the exponents by subtracting them:} \\ \frac{10^9}{10^1}=10^{9-1} \\ \frac{10^9}{10^1}=10^8 \end{gathered}\begin{gathered} \frac{9}{4.5}\times\text{ }\frac{10^9}{10^1}\text{ = 2 }\times10^8 \\  \\ \text{when dividing decimals, }the\text{ least number of }significant\text{ }figures\text{ in the problem}, \\ \text{ }\det ermines\text{ the significant figures in the answer} \\ 9\text{ = 1 significant, 4.5 = 2 significant} \\ \text{The least significant is 1} \\  \\ \frac{9\text{ }\times10^9}{4.5\text{ }\times10^1}\text{ = 2 }\times10^8 \end{gathered}

6 0
1 year ago
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