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3 years ago

Find the value of x, rounded to the nearest tenth.

Mathematics

1 answer:

V125BC [204]3 years ago

7 0

Answer:

x=6.2

Step-by-step explanation:

So to find this we can use the trignometric function sin(Θ):

Perpendicular = x

Hypotonuse = 18

Θ = 20°

sin(Θ) = $\frac{Perpendicular}{Hypotonuse}$

Substituting our values we get:

Sin(20°) = $\frac{x}{18}$

Using a calculator we get:

0.34202 = $\frac{x}{18}$

$x=0.34202*18$

x=6.15636

So rounding that to the nearest tenth :

x=6.2

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Step-by-step explanation:

$(x^{3} x^{-6} )^{2}$ is the expression given to be solved.

First of all let us have a look at 3 formulas:

$1.\ p^a \times p^b = p^{(a+b)}\\2.\ (p^a \times q^b)^c = (p^{a})^c \times (q^{b})^c\\3.\ (p^a)^b = p^{a\times b}$

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Applying formula (1):

$(x^{3} x^{-6} )^{2}$

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$p=x, a =3, b=-6$

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So, $(x^{3} x^{-6} )^{2}$ is reduced to $(x^{-3} )^{2}$

Applying formula (2):

Comparing the terms of $(x^{3} x^{-6} )^{2}$ with $(p^a \times q^b)^c$

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So, $(x^{3} x^{-6} )^{2}$ is reduced to $x^6 x^{-12}$ .

So, the answers can be:

$(x^{-3} )^{2}$

$x^6 x^{-12}$

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Step-by-step explanation:

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