A dilation by a scale factor of 6 will cause the radius to increase by a factor of 6. So
. If we plug this value of r into the formula for the circumference of a circle we get
So basically N = 12r. You haven't given me any original radius, so I can't give you a constant for N, but if you do have that original radius you can just plug that into 12r.
Based on your comment, N = 12 * 9 = 108 inches
let Leo's speed = x mile/hr
so Ethan speed = (x + 6) miles per hour
Distance = Speed × Time
so distance traveled by Leo in 2 hours = x × 2 = 2x
and distance traveled by Ethan in 1.5 hours = 1.5( x + 6 ) = 1.5x + 9
since they meet on the path , After Ethan has ridden 1.5 hours and Leo has ridden 2 hours , so together the have traveled on complete path that is 65 miles.
⇒ distance traveled by leo in 2 hours + distance traveled by Ethan in 1.5 hours = 65
⇒ 2x + 1.5x + 9 = 65
⇒ 3.5x = 56
⇒ x = 16
Hence Leo's speed = x miles/hr = 16 miles/hour
and Ethan's speed = (x + 6) miles/hr = 16 +6 = 22 mile/hour
Answer:
The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.
Step-by-step explanation:
According to DeMorgan's Theorem:
(W.X + Y.Z)'
(W.X)' . (Y.Z)'
(W'+X') . (Y' + Z')
Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.
For the original function:
(W . X + Y . Z)'
= (1 . 1 + 1 . 0)
= (1 + 0) = 1
For the compliment:
(W' + X') . (Y' + Z')
=(1' + 1') . (1' + 0')
=(0 + 0) . (0 + 1)
=0 . 1 = 0
Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.
Without the parenthesis the compliment equation looks like this:
W' + X' . Y' + Z'
1' + 1' . 1' + 0'
0 + 0 . 0 + 1
Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.
Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.
Answer:
a) 1+2+3+4+...+396+397+398+399=79800
b) 1+2+3+4+...+546+547+548+549=150975
c) 2+4+6+8+...+72+74+76+78=1560
Step-by-step explanation:
We know that a summation formula for the first n natural numbers:
1+2+3+...+(n-2)+(n-1)+n=\frac{n(n+1)}{2}
We use the formula, we get
a) 1+2+3+4+...+396+397+398+399=\frac{399·(399+1)}{2}=\frac{399· 400}{2}=399· 200=79800
b) 1+2+3+4+...+546+547+548+549=\frac{549·(549+1)}{2}=\frac{549· 550}{2}=549· 275=150975
c)2+4+6+8+...+72+74+76+78=S / ( :2)
1+2+3+4+...+36+37+38+39=S/2
\frac{39·(39+1)}{2}=S/2
\frac{39·40}{2}=S/2
39·40=S
1560=S
Therefore, we get
2+4+6+8+...+72+74+76+78=1560
