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Inessa [10]
3 years ago
12

Combine the radicals 5sqrt27-17sqrt3

Mathematics
2 answers:
Bess [88]3 years ago
6 0

Answer:

-2 sqrt3.

Step-by-step explanation:

sqrt27 = sqrt9 * sqrt3

= 3 sqrt3   So:

5sqrt27 - 17sqrt3

= 5*3 sqrt3 - 17 sqrt 3

= 15 sqrt 3 - 17 sqrt 3

= -2 sqrt3.

kow [346]3 years ago
4 0

For this case we must simplify the following expression:

5 \sqrt {27} -17 \sqrt {3}

We have to:

27 = 3 * 3 * 3 = 3 ^ 2 * 3

Substituting in the expression we have:

5 \sqrt {3 ^ 2 * 3} -17 \sqrt {3}

For properties of powers and roots we have:

\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}

So:

5 * 3 \sqrt {3} -17 \sqrt {3} =\\15 \sqrt {3} -17 \sqrt {3} =

We subtract taking into account that different signs are subtracted and the sign of the major is placed:

-2 \sqrt {3}

Answer:

-2 \sqrt {3}

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Suppose Upper F Superscript prime Baseline left-parenthesis x right-parenthesis equals 3 x Superscript 2 Baseline plus 7 and Upp
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It looks like you're given

<em>F'(x)</em> = 3<em>x</em>² + 7

and

<em>F</em> (0) = 5

and you're asked to find <em>F(b)</em> for the values of <em>b</em> in the list {0, 0.1, 0.2, 0.5, 2.0}.

The first is done for you, <em>F</em> (0) = 5.

For the remaining <em>b</em>, you can solve for <em>F(x)</em> exactly by using the fundamental theorem of calculus:

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On the other hand, if you're expected to <em>approximate</em> <em>F</em> at the given <em>b</em>, you can use the linear approximation to <em>F(x)</em> around <em>x</em> = 0, which is

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Then <em>F</em> (0) = 5, <em>F</em> (0.1) ≈ 5.7, <em>F</em> (0.2) ≈ 6.4, <em>F</em> (0.5) ≈ 8.5, and <em>F</em> (2.0) ≈ 19. Notice how the error gets larger the further away <em>b </em>gets from 0.

A <em>better</em> numerical method would be Euler's method. Given <em>F'(x)</em>, we iteratively use the linear approximation at successive points to get closer approximations to the actual values of <em>F(x)</em>.

Let <em>y(x)</em> = <em>F(x)</em>. Starting with <em>x</em>₀ = 0 and <em>y</em>₀ = <em>F(x</em>₀<em>)</em> = 5, we have

<em>x</em>₁ = <em>x</em>₀ + 0.1 = 0.1

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