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3 years ago

Please solve!!!!!!!!! Thanks!

Mathematics

1 answer:

Nat2105 [25]3 years ago

4 0

Answer:

a) 60°
b) 80°
c) 100°
d) 50°
e) 30°

Step-by-step explanation:

The key here is that AB ║ EC. This makes arc AE have the same measure as arc BC. Since those have the same measure as AB and the three arcs together make a semicircle, each has measure 180°/3 = 60°.

Then the various arc measures are:

AB = 60°
BC = 60°
CD = 80° (given)
DE = 100° . . . . . since CDE is 180°
EA = 60°

Then your answers are ...

a) AE = 60°

b) ∠ABD = (1/2)(DE +EA) = (1/2)(100° +60°) = 80°

c) ∠DFC = (1/2)(CD +EB) = (1/2)(80° + (60° +60°)) = 100°

d) ∠P = (1/2)(DA -AB) = (1/2)(100° +60° -60°) = 50°

e) ∠PAB = (1/2)(AB) = (1/2)(60°) = 30°

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What is 5.21 divided 0.7?

AURORKA [14]

Answer: 7.44285714 to be exact but its is around 7.44

Step-by-step explanation:

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3 years ago

A survey was conducted of 42 students who take math, Eng lish, or history. In the survey, 15 students said they take math and hi

strojnjashka [21]

Answer:

16 students take only English or only Math.

Step-by-step explanation:

We can solve this problem by treating these values as sets, and building the Venn Diagram.

I am going to say that:

A is the number of students who take Math.

B is the number of students who take English.

C is the number of students who take History.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a is the number of students that only take Math, $A \cap B$ is the number of students who take both Math and English, $A \cap C$ is the number of students that take both Math and History, and $A \cap B \cap C$ is the number of students that take all these classes.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

This diagram has the following subsets:

$a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

There were 42 students suveyed. This means that:

$a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 42$

We start finding the values from the intersection of three sets.

12 students said they take math, English, and history. This means that:

$A \cap B \cap C = 12$

18 students said they take English and history. This also takes into account those who take math, english and history. So:

$(B \cap C) + (A \cap B \cap C) = 18$

$B \cap C = 6$

17 students said they take math and English.

$(A \cap B) + (A \cap B \cap C) = 17$

$A \cap B = 5$

15 students said they take math and history

$(A \cap C) + (A \cap B \cap C) = 15$

$A \cap C = 3$

2 students said they only take history.

$c = 2$

How many students take only English or only math?

This is a + b, that we can find by the following formula:

$a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 42$

$a + b + 2 + 5 + 3 + 6 + 12 = 42$

$a + b = 16$

16 students take only English or only Math.

4 0

3 years ago

In ben's class, there were five exams given throughout the semester. if the average of his first four exams was a 90, what was h

balu736 [363]

Number of exams per semester = 5
Average of first four exams = 90

Therefore,
Total of first four exams = 90*4 = 360 marks

Based on 5 exams,

Average = 360/5 = 72

6 0

3 years ago

HELP! I need the answer ASAP

dalvyx [7]

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