A right triangle must satisfy this equation: a^2 + b^2 = c^2

a= shortest length
b=medium length
c=the longest length

a triangle with sides 4,5,6 will satisfy not this equation because:
4^2 + 5^2 ≠ 6^2
16+25 = 41

41 ≠ 36

so the answer is C.

Hope this helps.

-Jabba

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3 years ago

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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}.

masya89 [10]

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

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -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.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<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)

Taking exponents of both sides, we have

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

Replacing x with y, we have

$y = e^{x}$

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

Replacing y with h⁻¹(x), we have

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

Learn more about inverse image of a function here:

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2 years ago

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