26.2 miles using a fraction is 131/5
Answer: 10, 11, & 12
<u>Step-by-step explanation:</u>
Let x represent the age of the youngest child.
Their ages are consecutive so,
Youngest: x
Middle: x + 1
Oldest: x + 2
The age of the Youngest squared (x²) equals 8 times the Oldest [8(x + 2)] plus 4.
x² = 8(x + 2) + 4
x² = 8x + 16 + 4
x² = 8x + 20
x² - 8x - 20 = 0
(x - 10)(x + 2) = 0
x - 10 = 0 or x + 2 = 0
x = 10 or x = -2
Since age cannot be negative, x = -2 is not valid
So, the Youngest (x) is 10
the Middle (x + 1) is 11
and the Oldest (x + 2) is 12
One way to understand division is to look at it as repeated
subtraction. When you "divide by" a divisor number, you're
asking "how many times can I subtract this divisor from the
dividend, before the dividend is all used up ?".
Well, if the divisor is ' 1 ', then you're taking ' 1 ' away from the
dividend each time, and the number of times will be exactly
the same as the dividend.
If the divisor is more than ' 1 ', then you subtract more than ' 1 '
from the dividend each time, and the number of times you can
do that is less than the dividend itself.
If the divisor is less than ' 1 ', then you only take away a piece of
' 1 ' each time. You can do that more times than the number in
the dividend, because you only take away a piece each time.
Put your hands on your head and breath while doing that walk around slowly until you feel cooled down
F1 . . . 100% of it = 900N is in the +x direction.
F2 . . . 70.7% of it (cos45°, 530.3N) is in the +x direction,
and 70.7% of it (sin45°, 530.3N) is in the +y direction.
F3 . . . 80% of it (520N) is in the -x direction,
and 60% of it (390N) is in the +y direction.
Total x-component: 900 + 530.3 - 520 = 1,950.3 N
Total y-component: 530.3 + 390 = 920.3 N
Magnitude of the resultant = √ (x² + y²)
= √(1950.3² + 920.3²)
= √4,650,070.09
= 2,156.4 N .
Angle of the resultant, measured counterclockwise
from the +x axis, is
tan⁻¹ (y / x)
= tan⁻¹ (920.3 / 1950.3)
= tan⁻¹ (0.4719)
= about 25.3° .
Caution:
The same fatigue that degrades my ability to READ the question accurately
may also compromise the accuracy of my solutions. Before you use this
answer for anything, check it, check it, check it !
