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igor_vitrenko [27]
3 years ago
14

It is 11 a.m. in the Eastern Time Zone; therefore, what time is it in the Pacific Time Zone

Mathematics
1 answer:
lana66690 [7]3 years ago
6 0
Pacific time is 3 hours before Eastern time so subtract 3 hours from 11:00 am and you get 8:00 am
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Subtract:<br>4 4/9-2 7/9<br>​
defon

The answer is

1 2/3:

3 0
3 years ago
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Help please I don’t understand this problem
Molodets [167]

Answer:

A

Step-by-step explanation:

We know that identity:

cos^2x+sin^2x=1.

From that we have:

cos^2x=1-sin^2x.

So it is:

cos^2x-sin^x=(1-sin^2x)-sin^2x=1-2sin^2x.

And this is A).

B) is not treu because cos^2 is not sin^2x+1

C) is not true because cos^2x-sin^x=1 is false.

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3 years ago
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denpristay [2]

Answer:

in triangle PQR, <P - <Q = 20 degrees, <Q - <R = 50 degrees, find <P, <Q, <R

Q = 30, R = 80, P = 50

Step-by-step explanation:

in triangle PQR, <P - <Q = 20 degrees, <Q - <R = 50 degrees, find <P, <Q, <R

R  =  Q +50, 80=30+50

P  =  Q + 20, 50=30+20

Q = 30, R = 80, P = 50

3 0
2 years ago
6) A garden is 90m long and 50m wide. A path of 5m width is built outside and around it.
kompoz [17]
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5 0
3 years ago
Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.
frozen [14]

Note: Consider we need to find the vertices of the triangle A'B'C'

Given:

Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.

Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).

To find:

The vertices of the triangle A'B'C'.

Solution:

If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then

(x,y)\to (y,-x)

Using this rule, we get

A(-3,6)\to A'(6,3)

B(2,9)\to B'(9,-2)

C(1,1)\to C'(1,-1)

Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).

7 0
3 years ago
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