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allsm [11]
2 years ago
14

Which expression is equivalent to StartFraction StartRoot 2 EndRoot Over RootIndex 3 StartRoot 2 EndRoot EndFraction?.

Mathematics
1 answer:
miskamm [114]2 years ago
7 0

The expression is equivalent to \sqrt[6]{2}.

<h2>Given that</h2>

Expression; \dfrac{\sqrt{2}}{\sqrt[3]{2} }

<h3>We have to determine</h3>

The equivalent expression to the given expression.

<h3>According to the question</h3>

To determine the equivalent relation following all the steps given below.

Expression; \dfrac{\sqrt{2}}{\sqrt[3]{2} }

The equivalent expression is;

= \dfrac{\sqrt{2}}{\sqrt[3]{2} }\\\\= \dfrac{2^{\frac{1}{2}}}{2^{\frac{1}{3}}}\\\\ = 2^{\frac{1}{2}-\frac{1}{3}}\\\\ = 2 ^{\frac{3-2}{6}}\\\\= 2^{\frac{1}{6}}\\\\= \sqrt[6]{2}

Hence, the expression is equivalent to \sqrt[6]{2}.

To know more about Expression click the link given below.

brainly.com/question/16450385

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3 years ago
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Jason’s savings account has a balance of $2179. After 5 years , what will the amount of interest be at 6% compounded quarterly?
satela [25.4K]

Answer:

$755.80

Step-by-step explanation:

Determine the compound amount first and then subtract the principal from it, to find the amount of interest.

The compound amount formula is A = P (1 + r/n)^(nt), where

P is the initial principal, r is the interest rate as a decimal fraction, n is the number of compounding periods per year, and t is the number of years.  Here, P = $2179; t = 5 yrs; r = 0.06; and n = 4 (quarterly compounding).

We get:

A = $2179(1 + 0.06/4)^(4*5), or $2179(1.015)^20, or $2179(1.347) = $2937.80.

The compound amount is $2934.80.  Subtracting the $2179 principal results in the interest earned:  $755.80.

5 0
3 years ago
Explain how you I got 0.16
dybincka [34]
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4 0
3 years ago
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VikaD [51]

Answer:

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Step-by-step explanation

Answer:


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5 0
3 years ago
V = 1/3 bh for b the base of the cone
ch4aika [34]

Answer:

\huge\boxed{b=\dfrac{3V}{h}}

Step-by-step explanation:

V=\dfrac{1}{3}bh\qquad\text{multiply both sides by 3}\\\\3V=3\!\!\!\!\diagup^1\cdot\dfrac{1}{3\!\!\!\!\diagup_1}bh\\\\3V=bh\qquad\text{divide both sides by}\ h\neq0\\\\\dfrac{3V}{h}=\dfrac{bh}{h}\\\\\dfrac{3V}{h}=b\to \boxed{b=\dfrac{3V}{h}}

6 0
3 years ago
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