Mathematics, 27.06.2020 03:01, SethSimunek
What is the following product? RootIndex 3 StartRoot 16 x Superscript 7 Baseline EndRoot times RootIndex 3 StartRoot 12 x Superscript 9 Baseline EndRoot x squared (RootIndex 3 StartRoot 28 x squared EndRoot) x Superscript 5 Baseline (RootIndex 3 StartRoot 28 x EndRoot) 4 x squared (RootIndex 3 StartRoot 3 x squared EndRoot) 4 x Superscript 5 Baseline (RootIndex 3 StartRoot 3 x EndRoot)
Answers: 2
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }](/tpl/images/0695/6266/641d2.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9}](/tpl/images/0695/6266/ea79b.png)
![\sqrt[m]{a} * \sqrt[m]{b} = \sqrt[m]{a*b}](/tpl/images/0695/6266/cf5e0.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16x^7 * 12x^9}](/tpl/images/0695/6266/cf1ea.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16* x^7 * 12 * x^9}](/tpl/images/0695/6266/3e259.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^7 * x^9}](/tpl/images/0695/6266/862b3.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{7+9}}](/tpl/images/0695/6266/c5cb7.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{16}}](/tpl/images/0695/6266/0485d.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4*4*4*3* x^{16}}](/tpl/images/0695/6266/578d0.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4^3 *3* x^{16}}](/tpl/images/0695/6266/d7f16.png)
![a^{\frac{1}{m}} = \sqrt[m]{a}](/tpl/images/0695/6266/2aed2.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^3 *3* x^{16}})^{\frac{1}{3}}](/tpl/images/0695/6266/006ae.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^{3*{\frac{1}{3}}} *3^{{\frac{1}{3}}}* x^{16*{\frac{1}{3}}}})](/tpl/images/0695/6266/f1b98.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{\frac{16}{3}}}](/tpl/images/0695/6266/e036e.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5\frac{1}{3}}}](/tpl/images/0695/6266/91b1b.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5+\frac{1}{3}}}](/tpl/images/0695/6266/d94ed.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5}*{x ^\frac{1}{3}}}](/tpl/images/0695/6266/8b2c9.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 * x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}](/tpl/images/0695/6266/fc45e.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}](/tpl/images/0695/6266/1f460.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3} *\sqrt[3]{x} }](/tpl/images/0695/6266/27a4a.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3*x} }](/tpl/images/0695/6266/2117a.png)
![\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3x} }](/tpl/images/0695/6266/7bfe4.png)
Other questions on the subject: Mathematics
Mathematics, 21.06.2019 14:30, torresalysabeth
Find the value of 1715⋅211.though these numbers aren't quite as nice as the ones from the example or the previous part, the procedure is the same, so this is really no more difficult. give the numerator followed by the denominator, separated by a comma.
Answers: 1
Mathematics, 21.06.2019 17:30, emiller6462
Use the net as an aid to compute the surface area of the triangular prism. a) 550 m2 b) 614 m2 c) 670 m2 d) 790 m2
Answers: 1
Mathematics, 21.06.2019 23:40, hardwick744
Typically a point in a three dimensional cartesian coordinate system is represented by which of the following answers in the picture !
Answers: 1
Mathematics, 22.06.2019 01:30, QuestionsAnsweredNow
The actual length of side y is 22 cm. use the scale drawing to find the actual length of side x. a) 15.2 cm b) 17.1 cm c) 18.0 cm d) 28.3 cm
Answers: 1
Do you know the correct answer?
What is the following product? RootIndex 3 StartRoot 16 x Superscript 7 Baseline EndRoot times RootI...
Questions in other subjects:
English, 05.12.2021 18:40
Mathematics, 05.12.2021 18:40
Mathematics, 05.12.2021 18:40
Mathematics, 05.12.2021 18:40
English, 05.12.2021 18:50
Mathematics, 05.12.2021 18:50
History, 05.12.2021 18:50
