1) Fine the measure of the angle formed by the hands of a clock when it is 7:00.
2) For more precise measurements, a degree can be divided into 60 minutes, and each minute can be divided into 60 seconds. An angle measure of 42 degrees, 30 minutes, and 10 seconds is written at 42degrees30'10". ubtract this angle measure from the measure 81degrees21'15"
3)The coordinates of X are (a - 5, -2a). The coordinates of Y are (a + 1, 2a). If the distance between X and Y is 10, find the value of a.
2) For more precise measurements, a degree can be divided into 60 minutes, and each minute can be divided into 60 seconds. An angle measure of 42 degrees, 30 minutes, and 10 seconds is written at 42degrees30'10". ubtract this angle measure from the measure 81degrees21'15"
3)The coordinates of X are (a - 5, -2a). The coordinates of Y are (a + 1, 2a). If the distance between X and Y is 10, find the value of a.
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1. The numerals on a clock are 1, 2, ...,11, 12.
So the clock is divided into 12 equal parts.
If there are 360º in a clock (circle) then each hour piece is 360/12 or 30º.
From the 7 to the 12 we pass through 5 pieces. or 150.º
2. Remember how many seconds in a minute, and minutes in an hour.
81º21'15" - 42º30'10"
Start at the end as usual.
15" = 10" = 5"
21' - 30' will require us to barrow 60' from 1º
82' = 30 ' = 52'
And 80º (remember we barrowed 1)º -42º = 38º
Combine 38º52'5"
3. d = √[(x2 - x1)^2 + (y2 - y1)^2]
d = √[(a + 1)- (a - 5))^2 + ((2a ) - (-2a))^2
d = √[(6)^2 + (4a)^2]
d = √[(36) + (16a^2)]
d = √(16a^2 + 36)
Now 10 = ±√(16a^2 + 36)
100 = 16a^2 + 36
0 = 16a^2 - 64 = (4a + 8)(4a - 8)
So if (4a + 8) = 0 then a = -2
And if (4a - 8) = 0 then a = 2
So the clock is divided into 12 equal parts.
If there are 360º in a clock (circle) then each hour piece is 360/12 or 30º.
From the 7 to the 12 we pass through 5 pieces. or 150.º
2. Remember how many seconds in a minute, and minutes in an hour.
81º21'15" - 42º30'10"
Start at the end as usual.
15" = 10" = 5"
21' - 30' will require us to barrow 60' from 1º
82' = 30 ' = 52'
And 80º (remember we barrowed 1)º -42º = 38º
Combine 38º52'5"
3. d = √[(x2 - x1)^2 + (y2 - y1)^2]
d = √[(a + 1)- (a - 5))^2 + ((2a ) - (-2a))^2
d = √[(6)^2 + (4a)^2]
d = √[(36) + (16a^2)]
d = √(16a^2 + 36)
Now 10 = ±√(16a^2 + 36)
100 = 16a^2 + 36
0 = 16a^2 - 64 = (4a + 8)(4a - 8)
So if (4a + 8) = 0 then a = -2
And if (4a - 8) = 0 then a = 2
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1) a hand of a clock turn around equal 360,meanwhile a hour hand turn around equal 12 hours;
360/12=30(angle per hour),so 7:00, 30 multiply 7 equal 210, but on the other side we can find angle
360-210=150
2)42degrees30'10'' subtract 81degrees21'15'' which result must be negative,so we can use a skill,try 81degrees21'15'' subtract 42degrees30'10'' ,then add a negative; 15''-10''=5'' ; 21' subtract 30' need borrow 1 degrees(60minutes),21'+60'-30'=51' ; 81degrees - 42degrees - 1degrees(be borrowed)=38degrees,so the answer is negative 38degrees51'5''
3) square of distance = square of coordinates of ( y-x)
so you can find a equation 100 = square of 6 + square of 4a
the answer is a values 2 or -2
360/12=30(angle per hour),so 7:00, 30 multiply 7 equal 210, but on the other side we can find angle
360-210=150
2)42degrees30'10'' subtract 81degrees21'15'' which result must be negative,so we can use a skill,try 81degrees21'15'' subtract 42degrees30'10'' ,then add a negative; 15''-10''=5'' ; 21' subtract 30' need borrow 1 degrees(60minutes),21'+60'-30'=51' ; 81degrees - 42degrees - 1degrees(be borrowed)=38degrees,so the answer is negative 38degrees51'5''
3) square of distance = square of coordinates of ( y-x)
so you can find a equation 100 = square of 6 + square of 4a
the answer is a values 2 or -2