Difference between revisions of "1963 TMTA High School Algebra I Contest Problem 4"
Coolmath34 (talk | contribs) (Created page with "== Problem == The product of <math>3.5 \times 10^3</math> and <math>1.8 \times 10^4</math> is: <math>\text{(A)} 5.3 \times 10^7 \quad \text{(B)} \quad 6.3 \times 10^7 \quad \...") |
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The product of <math>3.5 \times 10^3</math> and <math>1.8 \times 10^4</math> is: | The product of <math>3.5 \times 10^3</math> and <math>1.8 \times 10^4</math> is: | ||
| − | <math>\text{(A)} 5.3 \times 10^7 \quad \text{(B)} \quad 6.3 \times 10^7 \quad \text{(C)} \quad 6.3 \times 10^{12} \quad \text{(D)} \quad | + | <math>\text{(A)} 5.3 \times 10^7 \quad \text{(B)} \quad 6.3 \times 10^7 \quad \text{(C)} \quad 6.3 \times 10^{12} \quad \text{(D)} \quad 5.3 \times 10^1 \quad \text{(E) NOTA}</math> |
== Solution == | == Solution == | ||
Latest revision as of 22:25, 1 February 2021
Problem
The product of 
and 
is:
Solution
Multiply:
![\[(3.5 \cdot 10^3)(1.8 \cdot 10^4) = (3.5 \cdot 1.8)(10^3 \cdot 10^4) = 6.3 \cdot 10^7\]](http://latex.artofproblemsolving.com/8/5/e/85e4b1497800b604c8a7304d0c2f9c7ffb42c948.png)
The answer is 
See Also
| 1963 TMTA High School Mathematics Contests (Problems) | ||
| Preceded by Problem 3 |
TMTA High School Mathematics Contest Past Problems/Solutions | Followed by Problem 5 |
