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mixas84 [53]
3 years ago
12

4. if m<AOB = 45° in circle O, what is m<ABC​

Mathematics
2 answers:
Tcecarenko [31]3 years ago
8 0

Answer:

m<ABC=22.5°

hope it helps

kap26 [50]3 years ago
5 0

Answer:

Answer:

m<ACB=22.5° [as inscribed angle is half to the central angle]

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What is the value of h^-1(-4)
scoundrel [369]

Answer:

-4/h

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Let Z be a standard normal variable. Find the value of z if z satisfies P(Z &gt; z) = 0.6331.
alisha [4.7K]

Answer:

z = -0.340

Step-by-step explanation:

We are given that Z is a standard normal variable.

We have to find the value of z such that

P(Z > z) = 0.6331

P( Z > z)=0.6331  

= 1 -P( Z \leq z)=0.6331  

=P( Z \leq z)=1 - 0.6331 = 0.3669  

Calculation the value from standard normal z table, we have,

P(z \leq -0.340) = 0.3669

Thus,

z = -0.340

4 0
3 years ago
Can you please help me <br> Step by step as well for brainliest
romanna [79]

Answer:

Step-by-step explanation:

42:7=ED:24

cross multiply 42 into 24.

Then divide by 7

you will get ED

Then use pythageorous theorem.

Ed square + 42 square=Ef square

Then do square root of Ef square

7 0
3 years ago
Solve each equation. Show all your work. Round your answers to four decimal places. (a) 7^4x=10,(b)ln(2)+In(4x-1)=5
KengaRu [80]
A.) 7^4x = 10
log base 10 (7^4X) = log base 10 (10)
4x log base 10 (7) = 1
4x (0.8451) = 1
3.3804x = 1
x = 0.2958

b.) ln(2) + ln(4x-1) = 5
ln (2 * 4x-1) = 5
ln (8x-2) = 5
log base (3) (8x-2) = 5
e^5 = 8x-2
e^5+2 = 8x
x = 18.8016
4 0
3 years ago
Which transformations could have taken place? select two options. r0, 90° r0, 180° r0, 270° r0, –90° r0, –270°
Lelechka [254]

The complete question is

"One vertex of a polygon is located at (3, –2). After a rotation, the vertex is located at (2, 3). Which transformations could have taken place? Select two options. R0, 90° R0, 180° R0, 270° R0, –90° R0, –270°"

The transformations that could have taken place are R0, 270°or R0, -90°

Let the vertex be V.

So, we have:

V = (3, –2)  before rotation

V' = (2, 3) after rotation

The possible transformations from V to V' are;

A 90 degrees counterclockwise rotation (r0, -90) is given :

(x, y) --(-y , x)

So,

(3, –2) --- (2, 3)

A 270 degrees clockwise rotation (r0, 270) is given :

(x, y) --(-y, x)

So,

(3, –2) --- (2, 3)

By comparing the endpoints of the rotations to V' = (2, 3), we can conclude that the transformations that could have taken place are: r0, 270°or r0, -90°.

Read more about transformations at:

brainly.com/question/11707700

#SPJ1

7 0
3 years ago
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