What you need to do here is a vector addition of the ship's velocity with respect to the water (V1), plus the velocity of the water with respect to land (V2).
Also remember that:
<span>i = unit vector east </span>
<span>j = unit vector north </span>
<span>V1 = 30 sin 192 i + 30 cos 192 j </span>
<span>V2 = 14 sin 112 i + 14 cos 112 j. </span>
The formula will look like this
<span>V1 + V2 = ?
Now please use the data to put the values into correct places and there you go! Hope this helps</span>
Answer:
1.
T mBAC = mB'A'C'
F 2mABC = mA'B'C'
F BC = 2B'C'
T 2XA = XA'
2
D'(-2/3; -1)
E'(-1;1)
F'(1;1)
G'(1;-1)
3
the centre is L(0;-2)
the scale factor is 4
length J'K' = 4JK
the measure of L is equal the measure of L'
<u>the</u><u> </u><u>table</u><u>:</u>
K(4;2) 4 4 16 16 0+16 -2+16 K'(16;14)
To find this, we have to find a value x such that 2*10^9 * x = 4*10^6. This is so that if we find x, 2*10^9 is x times smaller than 4*10^6. Then, using the multiplicative property of equality, we can divide both sides by 2*10^9 (to separate x) to get 4*10^6/2*10^9=x. Since 4/2=2 and 10^6/10^9=1/10^3 (since for every power of 10, we add 1, so we have 10*10*10*10*10*10/(10^6)*(10^3) and the 10^6’s therefore cross out), our answer is 2/10^3=2/1,000. However, this is clearly not the answer we’re looking for, and even if we find out how many times larger 2*10^9 is than 4*10^6 by switching the denominator and numerator, we end up with 500. Thus, you must have entered something wrong.
Feel free to ask further questions, and good luck!
Greetings!Arrange the equation in
slope y-intercept form to calculate the
rate of change (hourly rate):
Isolate for y:
Add -45x to both sides:
Divide both sides by -3:
The Equation Is:
The
Hourly Rate (Represented by the
slope) is:
$15
The
Signing Bonus (Represented by the
y-intercept) is:
$10
<em>The Graph for this Equation is attached below:
</em>
I hope this helped!
-Benjamin
