With a given parallel line and a given point on the line
we can use the point-line method: y-y0=m(x-x0)
where
y=mx+k is the given line, and
(x0,y0) is the given point.
Here
m=-10, k=-5, (x0,y0)=(-3,5)
=> the required line L is given by:
L: y-5=-10(x-(-3))
on simplification
L: y=-10x-30+5
L: y=-10x-25
1+2+3= 6 (even)
2+3+4= 9 (odd)
it can be both
Answer:
4400 is the answer
Step-by-step explanation: You move the decimal point to the right 3 spaces. If it was 10^-3 you would move the decimal point to the left 3 spaces.
Answer:
14+=+2L+%2B+2%28L-3%29 Simplify and solve for L
14+=+2L+%2B+2L+-+6 Combine like-terms.
14+=+4L-6 Add 6 to both sides.
20+=+4L Divide both sides by 4.
5+=+L The length is 5 meters.
W+=+L-3
W+=+5-3
W+=+2meters.
The width is 2 meters.
P+=+2L%2B2W
P+=+2%285%29%2B2%282%29
P+=+10%2B4
P+=+14meters.
Step-by-step explanation:
Answer:
a) The function is constantly increasing and is never decreasing
b) There is no local maximum or local minimum.
Step-by-step explanation:
To find the intervals of increasing and decreasing, we can start by finding the answers to part b, which is to find the local maximums and minimums. We do this by taking the derivatives of the equation.
f(x) = ln(x^4 + 27)
f'(x) = 1/(x^2 + 27)
Now we take the derivative and solve for zero to find the local max and mins.
f'(x) = 1/(x^2 + 27)
0 = 1/(x^2 + 27)
Since this function can never be equal to one, we know that there are no local maximums or minimums. This also lets us know that this function will constantly be increasing.
