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Given: circle k(O), m LM = 164°, m WK = 68°, m∠MLK = 65°. Find: m∠LMW

Musya8 [376]

Answer:

The measure of angle LMW is $m\angle LMW=67\°$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the measure of arc MW

we know that

The inscribed angle measures half that of the arc comprising

so

$m\angle MLK=\frac{1}{2}[arc\ MW+arc\ WK]$

substitute the given values

$65\°=\frac{1}{2}[arc\ MW+68\°]$

$130\°=[arc\ MW+68\°]$

$arc\ MW=130\°-68\°=62\°$

step 2

Find the measure of arc LK

we know that

$arc\ LM+arc\ MW+arc\ WK+arc\ LK=360\°$ -----> by complete circle

substitute the given values

$164\°+62\°+68\°+arc\ LK=360\°$

$294\°+arc\ LK=360\°$

$arc\ LK=360\°-294\°=66\°$

step 3

Find the measure of angle LMW

we know that

The inscribed angle measures half that of the arc comprising

so

$m\angle LMW=\frac{1}{2}[arc\ LK+arc\ WK]$

substitute the given values

$m\angle LMW=\frac{1}{2}[66\°+68\°]=67\°$

4 0

3 years ago

The number. Find the number.

ad-work [718]

Answer: 12

Step-by-step explanation:

18÷3=6

6+6=12

18+12=30

30-18=12

5 0

4 years ago

PLSSSS HELP ASSAP 50 POINTS FPR FIRST PERSON TO ANSWER

Dafna1 [17]

Answer:

a. 2/3 or 0.6666...

b. AB = 6cm and AD = 3cm

Step-by-step explanation:

28/42 = 2/3

14/21 = 2/3

42*1/7 = 6

21*1/7 = 3

6 0

4 years ago

PLEASE HELP!!!!!!!!

11Alexandr11 [23.1K]

Answer:

The answers would be:
7) y=-0.5x+2
8) y=-0.5x+2
9) y=0.6x-2
10) y=6x-4

Step-by-step explanation:

7 0

2 years ago

Use calculus to find the absolute maximum and minimum values of the function. (round all answers to three decimal places.) f(x)

Allisa [31]

Part A:

Given the function $f(x)=x+2\cos(x)$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6}$

Using the second derivative test,

$f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732$

Since the second derivative gives a negative number, the given function has a maximum point at $x=\frac{\pi}{6}$ .

And the maximum point is given by:

$f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}$

i.e. $\left(\frac{\pi}{6},\ 2.256\right)$

Part B:

Given the function $f(x)=e^{-x}-e^{-2x}$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931$

Using the second derivative test,

$f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5$

Since the second derivative gives a negative number, the given function has a maximum point at $x=0.6931$ .

And the maximum point is given by:

$f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}$

i.e. (0.693, 0.25)

3 0

3 years ago