Question

ma-cs-math 3 topic 4a exam fall 2025
answer form directions: use 50 pencil or marker. fill in the circle completely like this:
if changing an answer, erase cleanly, leaving no dark mark
building: carl sandburg high school (100)
course: math 3 (ma3101)
section: 09
teacher: ronquillo, edward
student: henderson, amiya naomi (272038)
free response
13 performance - score only 2pts
simplify the expression: (sqrt{108a^{13}b^9})
(teacher use only)
(○) 0 (○) .5 (○) 1 (○) 1.5 (○) 2
14 performance - score only 2pts
what is the simplest form of the expression? (sqrt{20} + sqrt{45} - sqrt{5})
(teacher use only)
(○) 0 (○) .5 (○) 1 (○) 1.5 (○) 2
15 performance - score only 2pts
rationalize and simplify the expression (\frac{2}{3 + sqrt{5}}).
(teacher use only)
(○) 0 (○) .5 (○) 1 (○) 1.5 (○) 2
16 performance - score only 2pts
describe the transformation of the parent function, (y = sqrt3{x}) for the function (y = -\frac{1}{2}sqrt3{x - 1} + 2).
(teacher use only)
(○) 0 (○) .5 (○) 1 (○) 1.5 (○) 2
17 performance - score only 2pts
solve the equation (-4x^3 = 32).
(teacher use only)
(○) 0 (○) .5 (○) 1 (○) 1.5 (○) 2
2025-12-09 | 16:41:15 (utc)
masterymanager.com
set: 36969496 | f

Explanation:

Question 13

Step1: Factor the radicand

Factor \(108a^{13}b^9\) into perfect - square factors and non - perfect - square factors. We know that \(108 = 36\times3=6^{2}\times3\), \(a^{13}=a^{12}\times a=(a^{6})^{2}\times a\), and \(b^{9}=b^{8}\times b=(b^{4})^{2}\times b\). So, \(\sqrt{108a^{13}b^{9}}=\sqrt{36\times3\times a^{12}\times a\times b^{8}\times b}\).

Step2: Use the property of square roots \(\sqrt{xy}=\sqrt{x}\cdot\sqrt{y}\) (\(x\geq0,y\geq0\))

\(\sqrt{36\times3\times a^{12}\times a\times b^{8}\times b}=\sqrt{36}\times\sqrt{a^{12}}\times\sqrt{b^{8}}\times\sqrt{3ab}\)

Step3: Simplify the perfect - square roots

We know that \(\sqrt{36} = 6\), \(\sqrt{a^{12}}=a^{6}\) (since \((a^{6})^{2}=a^{12}\)), and \(\sqrt{b^{8}}=b^{4}\) (since \((b^{4})^{2}=b^{8}\)). So, \(\sqrt{36}\times\sqrt{a^{12}}\times\sqrt{b^{8}}\times\sqrt{3ab}=6a^{6}b^{4}\sqrt{3ab}\)

Step1: Simplify each square root

Simplify \(\sqrt{20}\), \(\sqrt{45}\) and \(\sqrt{5}\). We know that \(\sqrt{20}=\sqrt{4\times5}=\sqrt{4}\times\sqrt{5}=2\sqrt{5}\), \(\sqrt{45}=\sqrt{9\times5}=\sqrt{9}\times\sqrt{5}=3\sqrt{5}\) and \(\sqrt{5}\) remains as it is.

Step2: Substitute the simplified forms into the expression

The expression \(\sqrt{20}+\sqrt{45}-\sqrt{5}\) becomes \(2\sqrt{5}+3\sqrt{5}-\sqrt{5}\)

Step3: Combine like terms

Since the terms are like radicals (they have the same radicand \(\sqrt{5}\)), we can combine the coefficients: \((2 + 3-1)\sqrt{5}=4\sqrt{5}\)

Step1: Rationalize the denominator

To rationalize the denominator of \(\frac{2}{3 + \sqrt{5}}\), we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(3+\sqrt{5}\) is \(3 - \sqrt{5}\). So, \(\frac{2}{3+\sqrt{5}}\times\frac{3 - \sqrt{5}}{3 - \sqrt{5}}=\frac{2(3-\sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}\)

Step2: Simplify the denominator

Using the difference of squares formula \((a + b)(a - b)=a^{2}-b^{2}\), where \(a = 3\) and \(b=\sqrt{5}\), the denominator \((3+\sqrt{5})(3 - \sqrt{5})=3^{2}-(\sqrt{5})^{2}=9 - 5 = 4\)

Step3: Simplify the numerator and the fraction

The numerator is \(2(3-\sqrt{5})=6 - 2\sqrt{5}\). Then the fraction becomes \(\frac{6 - 2\sqrt{5}}{4}\). We can factor out a 2 from the numerator: \(\frac{2(3-\sqrt{5})}{4}=\frac{3-\sqrt{5}}{2}\)

Answer:

\(6a^{6}b^{4}\sqrt{3ab}\)

Question 14