A parallel line has the same gradient
First you will need to rearrange the equation 10x+2y=-2
Then once you do that please comment
If you don’t know how please comment
B. End of the reconstruction.
The equation that matches the given situation is
. So, after two moves, Eric's elevation changed
meters above.
We know that in mathematics, subtraction means removing something from a group or a number of things. What is left in the group gets smaller when we subtract. The minuend is the first element we use. The subtrahend is the part that is being removed. The difference is the portion that remains after subtraction.
Assume that "negative" means climbing down and "positive" means climbing up. We must locate the elevation change in this area.
Given that Eric climbed straight down
meters. So we can write
.
Again Eric climbed straight up
meters. So, we can write
.
Then the change = 
=
=
=
=
=
=
=
Therefore, the equation that matches the given situation is
. So, after two moves, Eric's elevation changed
meters above.
Learn more about subtraction here -
brainly.com/question/24116578
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Answer: for a the answer is 140° and for b x=25°
Step-by-step explanation:
for a, when you have 2 parallel lines cut by a transversal,the corresponding angles are congruent. and the angle140 is congruent to the corresponding angle as they are vertical angles are congruent.
for b, angle x is congruent to the angle that is supplementary to 155 therefore 180-155=25°
Answer:
f + g)(x) = f (x) + g(x)
= [3x + 2] + [4 – 5x]
= 3x + 2 + 4 – 5x
= 3x – 5x + 2 + 4
= –2x + 6
(f – g)(x) = f (x) – g(x)
= [3x + 2] – [4 – 5x]
= 3x + 2 – 4 + 5x
= 3x + 5x + 2 – 4
= 8x – 2
(f × g)(x) = [f (x)][g(x)]
= (3x + 2)(4 – 5x)
= 12x + 8 – 15x2 – 10x
= –15x2 + 2x + 8
\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(
g
f
)(x)=
g(x)
f(x)
= \small{\dfrac{3x+2}{4-5x}}=
4−5x
3x+2
My answer is the neat listing of each of my results, clearly labelled as to which is which.
( f + g ) (x) = –2x + 6
( f – g ) (x) = 8x – 2
( f × g ) (x) = –15x2 + 2x + 8
\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}