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shusha [124]
3 years ago
11

What are the domain and range of the function f(x)=logx-5?

Mathematics
1 answer:
lora16 [44]3 years ago
7 0
<h2>Answer</h2>

Domain is (5, ∞), {x| x is greater than 5} for all integer n

Range is (-∞, ∞), {y| y belongs to Reals numbers}

<h2>Explanation</h2>

F(x) = log (x - 5)

Set up the equation for x

x - 5 > 0

or,

x > 5

So the domain is all values of x greater than 5 to infinity.

Range is set of all correct or valid values of y.

Range = (-∞, ∞), {y| y belongs to Reals numbers}


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Step-by-step explanation:

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Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL
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Answer:

KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47

Step-by-step explanation:

Given:

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1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

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m∠M = 50°

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\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}

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2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

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m\angle KLN=m\angle LNM=40^{\circ} (alternate interior angles)

m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ} (angles KNL and KLN are complementary).

So,

\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08

and

\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47

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