R= <span><span>√<span><span>x²</span>+<span>y²</span></span></span>
</span> = 16 or <span><span>(<span>x²</span>+<span>y²</span>)</span>
</span> = <span><span>16²</span>
</span>
You can solve for it, i believe, using the following formula
[(new-old)/old]*100
OR
(AmtIncreased/OriginalAmt)*100
Answer:
{x,y,z} = {-18,4,2}
Step-by-step explanation:
Solve equation [2] for the variable x
x = -10y + 2z + 18
Plug this in for variable x in equation [1]
(-10y+2z+18) + 9y + z = 20
- y + 3z = 2
Plug this in for variable x in equation [3]
3•(-10y+2z+18) + 27y + 2z = 58
- 3y + 8z = 4
Solve equation [1] for the variable y
y = 3z - 2
Plug this in for variable y in equation [3]
- 3•(3z-2) + 8z = 4
- z = -2
Solve equation [3] for the variable z
z = 2
By now we know this much :
x = -10y+2z+18
y = 3z-2
z = 2
Use the z value to solve for y
y = 3(2)-2 = 4
Use the y and z values to solve for x
x = -10(4)+2(2)+18 = -18
Answer:
D
Step-by-step explanation:
We need to graph both 
and 
and figure out <em>the x-value at which the graph of is LESS THAN that of the graph of .</em>
<em />
The attached picture shows the graph of in BLUE and the graph of in RED.
<em><u>We need to find x-value where the BLUE graph is "BELOW" the RED graph.</u></em> We can see that this occurs when x < 0.
Thus the correct answer is D
<span>The answer is B.No
Hope this helps </span>
