Please help me if you can...

For circle O, and m∠ABC = 55°. In the figure, ∠_____ and ∠____ have measures equal to 35°.

luda_lava [24]

Answer:

In the figure ∠ABO and ∠BCO have measures equal to 35°.

Step-by-step explanation:

<u><em>The complete question is</em></u>

For circle O, m CD=125° and m∠ABC = 55°

In the figure<____, (AOB, ABO, BOA) and <_____ (BCO, OBC,BOC) have measures equal to 35°

The picture in the attached figure

step 1

Find the measure of angle COB

we know that

$m\angle COB=arc\ CD$ ----> by central angle

we have

$arc\ CD=125^o$

therefore

$m\angle COB=125^o$

step 2

we know that

AB is a tangent to the circle O at point A

so

ABC and ABO are right triangles

In the right triangle ABC

Find the measure of angle BCA

Remember that

$m\angle BCA+m\angle\ ABC=90^o$ ---> by complementary angles in a right triangle

we have

$m\angle ABC=55^o$

substitute

$m\angle BCA+55^o=90^o$

$m\angle BCA=90^o-55^o=35^o\\$

step 3

In the triangle BCO

Find the measure of angle CBO

we know that

$m\angle CBO+m\angle COB+m\angle BCO=180^o$ ---> the sum of the interior angles in any triangle must be equal to 180 degrees

we have

$m\angle COB=125^o$

$m\angle BCO=m\angle BCA=35^o$ -----> have measure equal to 35 degrees

substitute

$m\angle CBO+125^o+35^o=180^o$

$m\angle CBO=180^o-160^o=20^o$

step 4

Find the measure of angle ABO

In the right triangle ABO

we know that

$m\angle ABC=m\angle CBO+m\angle ABO$ ----> by angle addition postulate

we have

$m\angle ABC=55^o$

$m\angle CBO=20^o$

substitute

$55^o=20^o+m\angle ABO$

$m\angle ABO=55^o-20^o=35^o$ ----> have measure equal to 35 degrees

therefore

In the figure ∠ABO and ∠BCO have measures equal to 35°.

3 0

2 years ago

Emily is buying new cold-weather gear for her family. She pays $100 for 5 pairs

user100 [1]

Answer:

500, or 400 if its 4

Step-by-step explanation:

5 0

2 years ago

Find the geometric mean of 242 and 8

Gemiola [76]

The formula to solve these is the square root of X x Y, So the problem would be 242 * 8 which = 1,936. Then you find the square root of 1,936 which = 44 . So that is your answer :) Hope it helps

8 0

2 years ago

Read 2 more answers

How do I factorise 35x+55

densk [106]

Find what is common between 35 and 55....and that would be 5....so factor 5 out

35x + 55 =
5(7x + 11) <==

5 0

2 years ago

Read 2 more answers

3.12 Speeding on the I-5, Part I. The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in Ca

marusya05 [52]

Answer:

a) 93.943% = 93.9%

b) 93.528% = 93.5%

c) Speed of the fastest 5% ≥ 80.5 miles/hour

d) 29.46% = 29.5%

Step-by-step explanation:

Mean, xbar = 72.6 miles/hour.

standard deviation, σ = 4.78 miles/hour

For each of the questions, we'll need to normalize the speeds.

a) The standardized score for 80 miles/hour is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ = (80 - 72.6)/4.78 = 1.55

To determine the probability of a car having speed less than 80 miles/hour, P(x < 80) = P(z < 1.55)

We'll use data from the normal probability table for these probabilities

P(x < 80) = P(z < 1.55) = 1 - P(z ≥ 1.55) = 1 - P(z ≤ -1.55) = 1 - 0.06057 = 0.93943

b) percent of passenger vehicles travel between 60 and 80 miles/hour.

60 miles/hour standardized = (60 - 72.6)/4.78 = -2.64

We'll use data from the normal probability table for these probabilities

P(60 < x < 80) = P(-2.64 < z < 1.55) = P(z ≤ 1.55) - P(z ≤ -2.64) = 0.93943 - 0.00415 = 0.93528

c) How fast to do the fastest 5% of passenger vehicles travel?

We'll use data from the normal probability table for these probabilities

Top 5% corresponds to a z-score of 1.65. P(z ≥ 1.65) = 0.95053

1.65 = (x - 72.6)/4.78

x = 80.487 miles/hour = 80.5 miles/hour.

d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.

70 miles/hour, standardized = (70 - 72.6)/4.78 = 0.54

P(x > 70) = P(z > 0.54) = 1 - P(z ≤ 0.54) = 1 - 0.7054 = 0.2946.

Hope this helps!!!!

6 0

2 years ago