Simplify
−
3
(
j
+
3
)
+
9
j
-
3
(
j
+
3
)
+
9
j
.
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Simplify each term.
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−
3
j
−
9
+
9
j
<
−
15
-
3
j
-
9
+
9
j
<
-
15
Add
−
3
j
-
3
j
and
9
j
9
j
.
6
j
−
9
<
−
15
6
j
-
9
<
-
15
Move all terms not containing
j
j
to the right side of the inequality.
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Add
9
9
to both sides of the inequality.
6
j
<
−
15
+
9
6
j
<
-
15
+
9
Add
−
15
-
15
and
9
9
.
6
j
<
−
6
6
j
<
-
6
Divide each term in
6
j
<
−
6
6
j
<
-
6
by
6
6
and simplify.
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Divide each term in
6
j
<
−
6
6
j
<
-
6
by
6
6
.
6
j
6
<
−
6
6
6
j
6
<
-
6
6
Simplify the left side.
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j
<
−
6
6
j
<
-
6
6
Simplify the right side.
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j
<
−
1
j
<
-
1
The result can be shown in multiple forms.
Inequality Form:
j
<
−
1
j
<
-
1
Interval Notation:
(
−
∞
,
−
1
)
Step-by-step explanation:
I cannot give an exact result but I will try.
What it means by this is to continue the graph up to 2010.
Based on its trajectory, you would simply draw the line on.
If I made a prediction on what it would end up on in 2010, I would say 50 percent.
We have to find the lengths of the diagonals KM and JL:
d ( KM ) = √ (( - a - b )² + ( 0 - c )²) = √ (( a + b )² + c² )
d ( JL ) = √ ( ( a - ( - b ) )² + ( 0 - c )²) = √ ( ( a + b )² + c² )
So the lengths of the diagonals KM and JL are congruent.
The lengths of the diagonals of the isosceles trapezoid are congruent.
Answer:
The x-coordinate of the point changing at ¼cm/s
Step-by-step explanation:
Given
y = √(3 + x³)
Point (1,2)
Increment Rate = dy/dt = 3cm/s
To calculate how fast is the x-coordinate of the point changing at that instant?
First, we calculate dy/dx
if y = √(3 + x³)
dy/dx = 3x²/(2√(3 + x³))
At (x,y) = (1,2)
dy/dx = 3(1)²/(2√(3 + 1³))
dy/dx = 3/2√4
dy/dx = 3/(2*2)
dy/dx = ¾
Then we calculate dx/dt
dx/dt = dy/dt ÷ dy/dx
Where dy/dx = ¾ and dy/dt = 3
dx/dt = ¾ ÷ 3
dx/dt = ¾ * ⅓
dx/dt = ¼cm/s
The x-coordinate of the point changing at ¼cm/s
Slope = (6 - 3)/(-2-2)
= 3/-4
= -3/4
answer
slope = -3/4
