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zimovet [89]
3 years ago
8

What is 64 thousandths in decimal form​

Mathematics
2 answers:
shusha [124]3 years ago
8 0
I am pretty sure it is 0.064. Hope this helps
Sedbober [7]3 years ago
8 0

The answer is <u>.00064</u>, because if you were to make in <u>Thousand </u>form it would look like this ➡️ 64,000. See the <u>3 Zeros</u>? that is what makes it a thousand. now wht woul happen if we switched that around? well it would look like this ➡️ .00064. as you can see, there are<u> 3 zeros</u>, like in the last example. so your answer is ''.00064''

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Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.
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a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is g^{-1}(x) = e^{x}

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

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⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

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x = ∛(y/3)

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Replacing y with f⁻¹(x), we have

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<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is g^{-1}(x) = e^{x}

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y =  ln(x)

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e^{y} = e^{lnx} \\e^{y} = x

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y = e^{x}

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is g^{-1}(x) = e^{x}

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

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Learn more about inverse image of a function here:

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