175% of 14.2 is saying that
1.75(14.2)=24.9
Any questions please just ask. Thank you.
Graph 1 is related to table C because the first 3 values are increasing, and the 4th value decreases. Or because at 1PM, 3PM, and 5PM, the temperature was increasing, but at 7PM the temperature decreased. Graph 1 shows the first 3 points increasing, and then decreasing at the 4th point.
Graph 2 is related to table A because as the time increases/goes on, the temperature decreases exponentially/continues to decrease at a higher rate than before. From 1-3PM the temperate decreases by 2°F, from 3-5PM it decreased by 8°F, from 5-7PM the temperature decreased by 17°F.
Graph 3 is related to table B because as the time increases/goes on, the temperature decreases at a steady rate of 1°F every 2 hours.
Answer:
Let the vectors be
a = [0, 1, 2] and
b = [1, -2, 3]
( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.
Let the cross product be another vector c.
To find the cross product (c) of a and b, we have
![\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C0%261%262%5C%5C1%26-2%263%5Cend%7Barray%7D%5Cright%5D)
c = i(3 + 4) - j(0 - 2) + k(0 - 1)
c = 7i + 2j - k
c = [7, 2, -1]
( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:
c / | c |
Where | c | = √ (7)² + (2)² + (-1)² = 3√6
Therefore, the unit vector is
![\frac{[7,2,-1]}{3\sqrt{6} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B7%2C2%2C-1%5D%7D%7B3%5Csqrt%7B6%7D%20%7D)
or
[ 
, 
, 
]
The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:
[ 
, 
, 
]
In conclusion, the two unit vectors are;
[ , , ]
and
[ , , ]
<em>Hope this helps!</em>
It means <span>9 is not an element of T. hope it helped</span>
The answer is C.
You have to subtract 13 from both sides.
