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lukranit [14]
3 years ago
7

Rita is spending more time at home to study and practice math. Her efforts are finally paying off. On her first assessment she s

cored 58 points, then she scores 63 and 68 on her next two assessments. If her scores continued to increase at the same rate, on which assessments will she be scoring above 85? Select all that apply.
Mathematics
1 answer:
sammy [17]3 years ago
7 0
<span>On the first assessment Rita scored 58 points. For each next assessment the score increases by 5: 58 + 6 = 63, 63 + 5 = 68, 68 + 5 = 72, etc. So a1 = 58, d = 5. This is an example of the arythmetic sequence: a n = a 1 + ( n - 1 ) * d ; We have to find n such that: 58 + ( n - 1 ) * 5 > 85. ( n - 1 ) * 5 > 27; n - 1 > 5.4; n > 6.4. Finally: n = 7. Answer: Rita will be scoring above 85 on the 7th and each next assessment. </span>
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Answer:

(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{15}

Step-by-step explanation:

Given

(\frac{6}{10})^3 * (\frac{5}{9})^2

Required

Solve:

(\frac{6}{10})^3 * (\frac{5}{9})^2

Simplify 6/10

(\frac{6}{10})^3 * (\frac{5}{9})^2 = (\frac{3}{5})^3 * (\frac{5}{9})^2

Express 9 as 3^2

(\frac{6}{10})^3 * (\frac{5}{9})^2 = (\frac{3}{5})^3 * (\frac{5}{3^2})^2

Apply law of indices

(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3}{5^3}* \frac{5^2}{(3^2)^2}

(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3}{5^3}* \frac{5^2}{3^4}

Express as a sing;e fraction

(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3*5^2}{5^3*3^4}

Apply law of indices:

(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5^{3-2}*3^{4-3}}

(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5^1*3^1}

(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5*3}

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3 years ago
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Answer:

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