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3 years ago

Two lines intersect. Find the values of b and c.

Mathematics

2 answers:

aliina [53]3 years ago

5 0

Answer:

b is 42 and c is 138............

Mandarinka [93]3 years ago

3 0

Step-by-step explanation:

I think b is 42 as well cause there are vertically opposite

ummm ....and c is (180- 42 because they are adjacent angles)

therefore b=42

c= 138

that is what I yhink sorry if it ain't the right answer

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Step-by-step explanation

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3 years ago

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makkiz [27]

Answer:

Step-by-step explanation:

Since the difference in each term number is 5, we put 5 in the place of the star.

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2 years ago

Two sections of a class took the same quiz. Section A had 15 students who had a mean score of 80, and Section B had 20 students

Nezavi [6.7K]

Answer: A. 85.7

Step-by-step explanation:

Given : Two sections of a class took the same quiz.

Section A had 15 students who had a mean score of 80, and Section B had 20 students who had a mean score of 90.

We know that , $\text{Mean}=\dfrac{\text{Sum of observations}}{\text{No. of observations}}$

Then , for section A :

$\text{Mean score }=\dfrac{\text{Sum of scores in sec A}}{\text{No. of students}}$

$\Rightarrow\ 80=\dfrac{\text{Sum of scores in sec A}}{15}\\\\\Rightarrow\ \text{Sum of scores in sec A}=80\times15=1200$

Similarly in Section B, $\text{Sum of scores in sec B}=90\times20=1800$

Total scores = Sum of scores in sec A+Sum of scores in sec B

=1200+1800=3000

Total students = Students in sec A +Students in sec B

=15+20=35

Now , the mean score for all of the students on the quiz = $\dfrac{\text{Total score}}{\text{Total students}}$

$=\dfrac{3000}{35}=85.7142857143\approx85.7$

Hence, the approximate mean score for all of the students on the quiz = 85.7

Thus , the correct answer is option A. 85.7.

7 0

3 years ago

HELPPP!!!! Both questions are incredibly confusing to me !!!!!

Darya [45]

Answer:

Step-by-step explanation:

I'm going to start with problem 3. You need to become familar with the kind of tricks teachers play on you.

Problem 3 depends on getting RS = RW.

RT = RT Reflexive property

<STR = <WTR A straight line having 1 rt angle actually has 2

ST = TU They are marked as equal

Triangle STR=Triangle WTR SAS

RS = RW Corresponding parts of = triangles are =

8x = 6x + 5 Subtract 6x from both sides

8x -6x = 6x - 6x + 5 Combine

2x = 5 Divide by 2

2x/2 = 5/2

x = 2.5

RU = 6*2.5 + 5

RU = 15 + 5

RU = 20

Now we can play with Question 4.

This question depends on the method used in three, although not entirely.

What you need to know is that W is on RT when you take a ruler and make RT longer. You can put W anywhere as long as it is on RT when it is made longer.

Directions

Make RT longer. Draw down and to your right.

Put a point anywhere on the length starting at T. Label this new point as W. There's your W. It goes anywhere on the part of RT made longer.

Draw UW.

Label UW as 8

Now draw another line from S to W. Guess what? By the methods used in question 3, it's also 8. So SW = 8

TW = TW Reflexive

<UTW = STW Same reason as in 3. UtW is a right angle

UT = ST Given (the marking tells you so.

ΔUTW = ΔSTW SAS

UW = SW Corresponding parts of = triangles are =

SW = 8

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2 years ago