Answer: About 83915 grams
.............................
Answer:
a reflection over the x-axis and then a 90 degree clockwise rotation about the origin
Step-by-step explanation:
Lets suppose triangle JKL has the vertices on the points as follows:
J: (-1,0)
K: (0,0)
L: (0,1)
This gives us a triangle in the second quadrant with the 90 degrees corner on the origin. It says that this is then transformed by performing a 90 degree clockwise rotation about the origin and then a reflection over the y-axis. If we rotate it 90 degrees clockwise we end up with:
J: (0,1) , K: (0,0), L: (1,0)
Then we reflect it across the y-axis and get:
J: (0,1), K:(0,0), L: (-1,0)
Now we go through each answer and look for the one that ends up in the second quadrant;
If we do a reflection over the y-axis and then a 90 degree clockwise rotation about the origin we end up in the fourth quadrant.
If we do a reflection over the x-axis and then a 90 degree counterclockwise rotation about the origin we also end up in the fourth quadrant.
If we do a reflection over the x-axis and then a reflection over the y-axis we also end up in the fourth quadrant.
The third answer is the only one that yields a transformation which leads back to the original position.
Answer:
x=16/3
Step-by-step explanation:
5x−2(x+1)=14
Distribute
5x -2x -2 =14
Combine like terms
3x -2 =14
Add 2 to each side
3x-2+2 =14+2
3x=16
Divide each side by 3
3x/3 = 16/3
x=16/3
Answer:
Step-by-step explanation:
x
2
+
x
−
6
=
(
x
+
3
)
(
x
−
2
)
x
2
−
3
x
−
4
=
(
x
−
4
)
(
x
+
1
)
Each of the linear factors occurs precisely once, so the sign of the given rational expression will change at each of the points where one of the linear factors is zero. That is at:
x
=
−
3
,
−
1
,
2
,
4
Note that when
x
is large, the
x
2
terms will dominate the values of the numerator and denominator, making both positive.
Hence the sign of the value of the rational expression in each of the intervals
(
−
∞
,
−
3
)
,
(
−
3
,
−
1
)
,
(
−
1
,
2
)
,
(
2
,
4
)
and
(
4
,
∞
)
follows the pattern
+
−
+
−
+
. Hence the intervals
(
−
3
,
−
1
)
and
(
2
,
4
)
are both part of the solution set.
When
x
=
−
1
or
x
=
4
, the denominator is zero so the rational expression is undefined. Since the numerator is non-zero at those values, the function will have vertical asymptotes at those points (and not satisfy the inequality).
When
x
=
−
3
or
x
=
2
, the numerator is zero and the denominator is non-zero. So the function will be zero and satisfy the inequality at those points.
Hence the solution is:
x
∈
[
−
3
,
−
1
)
∪
[
2
,
4
)
graph{(x^2+x-6)/(x^2-3x-4) [-10, 10, -5, 5]}
The digits in the ten-thousands place is 10,000 times the value of a digit, right? For example, 10,000 is 10,000 times 1, and one is a mere digit. The thousands place follows the same rule, with 1,000 being 1,000 times 1. Ergo, when compared, you could think of it as 10,000/1,000 = 10. We can think of this as a digit in the ten-thousands place is 10 times the value of the same digit in the thousands place.
