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Semmy [17]
3 years ago
12

In the circle, O is the centre and AB is a diameter.

Mathematics
1 answer:
Burka [1]3 years ago
7 0

Answer:

Step-by-step explanation:

Answer:

a). m∠a = 62°

b). m∠b = 118°

Step-by-step explanation:

a). Since, an angle inscribed in a semicircle is a right angle,

  m∠ACB = 90°

  By triangle sum theorem in ΔABC,

  m∠CAB + m∠ACB + m∠ABC = 180°

  28° + 90° + m∠ABC = 180°

  118° + m∠ABC = 180°

  m∠ABC = 180°- 118°

  a = 62°

b). Since, quadrilateral ABCD is a cyclic quadrilateral,

   m∠ADC + m∠ABC = 180°

   b + a = 180°

   b + 62° = 180°

   b = 180° - 62°

   b = 118°

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Here's link to the answer:

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4 0
2 years ago
Identify the slope and y-intercept of the graph of the equation.
Ksivusya [100]

Answer:

B

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
25 POINTS
uysha [10]

Answer:

a reflection over the x-axis and then a 90 degree clockwise rotation about the origin

Step-by-step explanation:

Lets suppose triangle JKL has the vertices on the points as follows:

J: (-1,0)

K: (0,0)

L: (0,1)

This gives us a triangle in the second quadrant with the 90 degrees corner on the origin. It says that this is then transformed by performing a 90 degree clockwise rotation about the origin and then a reflection over the y-axis. If we rotate it 90 degrees clockwise we end up with:

J: (0,1) , K: (0,0), L: (1,0)

Then we reflect it across the y-axis and get:

J: (0,1), K:(0,0), L: (-1,0)


Now we go through each answer and look for the one that ends up in the second quadrant;

If we do a reflection over the y-axis and then a 90 degree clockwise rotation about the origin we end up in the fourth quadrant.

If we do a reflection over the x-axis and then a 90 degree counterclockwise rotation about the origin we also end up in the fourth quadrant.

If we do a reflection over the x-axis and then a reflection over the y-axis we also end up in the fourth quadrant.

The third answer is the only one that yields a transformation which leads back to the original position.

4 0
2 years ago
Read 2 more answers
30 years ago zoe was 2/3 as old as luke, 18 years ago zoe was 5/6 as old as luke how old are they now
barxatty [35]

Based on the mathematical statements, Zoe is 56 years old and Luke is 66 years old, now

<h3>How to determine how old they are now?</h3>

From the question, we have the following statements that can be used in our computation:

<u>30 years ago</u>

Zoe was 2/3 as old as Luke

<u>18 years ago</u>

Zoe was 5/6 as old as Luke

Let their present ages be represented as

Zoe = x

Luke = y

So, we have the following representations

<u>30 years ago</u>

Zoe was 2/3 as old as Luke

x - 36 = 2/3(y - 36)

<u>18 years ago</u>

Zoe was 5/6 as old as Luke

x - 18 = 5/6(y - 18)

So, we have the following system of equations

x - 36 = 2/3(y - 36)

x - 18 = 5/6(y - 18)

Make x the subject in x - 18 = 5/6(y - 18)

x = 5/6(y - 18) + 16

Substitute x = 5/6(y - 18) + 16 in x - 36 = 2/3(y - 36)

5/6(y - 18) + 16 - 36 = 2/3(y - 36)

Open the brackets

5/6y - 15 + 16 - 36 = 2/3y - 24

Evaluate the like terms

5/6y - 35 = 2/3y - 24

Multiply through by 6

5y - 210 = 4y - 144

Evaluate the like terms

y = 66

Substitute y = 66 in x = 5/6(y - 18) + 16

x = 5/6(66 - 18) + 16

Evaluate

x = 56

Recall that

Zoe = x

Luke = y

So, we have

Zoe = x = 56

Luke = y = 66

Hence, they are 56 and 66 years, now

Read more about equations at

brainly.com/question/2476251

#SPJ1

5 0
9 months ago
The recursive rule for a sequence is shown.an=an−1+9a1=21 what is the explicit rule for this sequence?
bixtya [17]

Answer:

Meaning of Explicit rule , is the general rule for a sequence.

The given sequence is defined as,

a_{n}=a_{n-1}+9a_{1}-21----(1)

Common Difference(d) of an airthmetic Sequence= N th term - (N-1) th term

                                 =a_{n}-a_{n-1}\\\\=a_{n-1}-a_{n-2}                                            

Equation 1 can be written as,

a_{n}-a_{n-1}=9 a_{1}-21\\\\ d =9a_{1}-21

General rule or explicit rule for this sequence is given as:

a_{n}=a_{1} +(n-1)d\\\\a_{n}=a_{1}+(n-1)(9a_{1}-21)\\\\a_{n}=a_{1}+9na_{1}-21 n-9a_{1}+21\\\\a_{n}=-8a_{1}+21+n(9a_{1}-21)

6 0
3 years ago
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