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Sphinxa [80]
3 years ago
15

Please help! The answer isn’t 36! Find the value of the expression:

Mathematics
1 answer:
Brums [2.3K]3 years ago
8 0

Answer:

3.6

Step-by-step explanation:

Hi there!

We are given this expression:

.6√36 (.6*√36)

And we want to find the value of it

.6 can be re-written as 0.6

In that case,

0.6*√36

First, simplify what's under the radical: √36, which is equal to 6 (6*6=36)

The expression then becomes:

0.6*6

Multiply those numbers together

0.6*6=<u>3.6</u>

Hope this helps!

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Which choice is equivalent to the product below when x&gt;0?
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Answer:

D

Step-by-step explanation:

Using the rule of radicals

\sqrt{\frac{a}{b} } = \frac{\sqrt{a} }{\sqrt{b} }

\sqrt{a\\} × \sqrt{b} ⇔ \sqrt{ab}

Given

\sqrt{\frac{6}{x} } × \sqrt{\frac{x^2}{24} }

= \frac{\sqrt{6} }{\sqrt{x} } × \frac{x^2}{24}

= \frac{\sqrt{6} }{\sqrt{x} } × \frac{x}{2 \sqrt{6\\} }

Cancel \sqrt{6} on numerator/ denominator

= \frac{1}{\sqrt{x} } × \frac{x}{2\\}

= \frac{1}{\sqrt{x} } × \frac{(\sqrt{x})^2 }{2}

Cancel \sqrt{x} on numerator/ denominator, leaving

= \frac{\sqrt{x} }{2} → D

4 0
3 years ago
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Rufina [12.5K]

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6 0
3 years ago
2x =56<br> 2x = 58<br> 2x = 204<br> 2x = 304
Lady_Fox [76]
.........
....................
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3 years ago
Using Newton's Method, what is the formula for determining the nth term in a sequence of approximations to a solution of a polyn
lorasvet [3.4K]

Step-by-step explanation:

in newton rapson method

x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}

Let n=n-1

x_{(n-1)+1}=x_{n-1}-\frac{f\left(x_{n-1}\right)}{f^{\prime}\left(x_{n-1}\right)}

x_{n}=x_{n-1}-\frac{f\left(x_{n-1}\right)}{f^{\prime}\left(x_{n-1}\right)}

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Svetllana [295]
7.5
f(1) = 120
f(2) = -60
f(3) = 30
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f(5) = 7.5
6 0
3 years ago
Read 2 more answers
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