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1 year ago

34. Find the average rate of change for x 0.5 to 35 for the following temperatures in degrees Fahrenheit

Mathematics

1 answer:

earnstyle [38]1 year ago

3 0

The average rate of change between the interval x = 0.5 to x = 3.5 is: 5.

x = 3.5 represents 3.5 hours after noon.

<h3>How to Find Average Rate of Change Over an Interval</h3>

Given an interval of x = a to x = b, the average rate of change over the given interval of a function can be calculated using the formula below:

$\frac{f(b) - f(a)}{b - a}$

Given table representing the function as shown in the image attached below, let's determine the average rate of change for x = 0.5 to x = 3.5.

a = 0.5

f(a) = 47 (from the table, when x = 0.5, y = 47)

b = 3.5

f(b) = 62 (from the table, when x = 3.5, y = 62)

Plug the values into $\frac{f(b) - f(a)}{b - a}$

Average rate of change = $\frac{62 - 47}{3.5 - 0.5} = \frac{15}{3} = 5$

Average rate of change = 5

x = 3.5 represents 3.5 hours after noon.

Learn more about average rate of change on:

brainly.com/question/11627203

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pantera1 [17]

6x-5y=17
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2 years ago

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

Mice21 [21]

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:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

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

LM = 45

m∠M = 50°

So,

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

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

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

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (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

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