Let
x-------> the whole number
y------> the decimal number
we know that
------> equation 1
assume the value of x and then calculate the value of y
Let
x=280
solve for y in the equation 1
substitute the value of x
therefore
<u>the expression is equal to</u>
280*0.001=0.28
<h3>Possible coordinates of R = -5 or 5</h3><h3>Possible coordinates of T = -7 or 7</h3>
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Explanation:
If we're on a number line, then R could be at either R = 5 or R = -5. This is so the distance from A to R is 5 units. Distance is never negative. You count out the spaces to get the distance, or use subtraction and absolute value.
Saying "distance from A to R is 5" can be written as AR = 5. Meaning segment AR is 5 units long.
Now if AT = 7, then T could be at 7 or -7 on the number line. The reasoning as similar as to why R could be at -5 or 5.
Well, not specific. But after 7 hour the bus will be 210mi away
The intersection line of two planes is the cross product of the normal vectors of the two planes.
p1: z=4x-y-13 => 4x-y-z=13
p2: z=6x+5y-13 => 6x+5y-z=13
The corresponding normal vectors are:
n1=<4,-1,-1>
n2=<6,5,-1>
The direction vector of the intersection line is the cross product of the two normals,
vl=
i j k
4 -1 -1
6 5 -1
=<1+5, -6+4, 20+6>
=<6,-2,26>
We simplify the vector by reducing its length by half, i.e.
vl=<3,-1,13>
To find the equation of the line, we need to find a point on the intersection line.
Equate z: 4x-y-13=6x+5y-13 => 2x+6y=0 => x+3y=0.
If x=0, then y=0, z=-13 => line passes through (0,0,-13)
Proceed to find the equation of the line:
L: (0,0,-13)+t(3,-1,13)
Convert to symmetric form:
=>
The angle VSR is a 100° angle
