Answer:
no, because intergers are negative. they will never be able to be greater
Step-by-step explanation:
The value of k is 2 if the polynomial x³ - 3x² - 10x + 24 is divided by the expression (x - 2).
<h3>What is synthetic division?</h3>
In the case of dividing by a linear factor, the synthetic division is a shorthand, or quicker, a technique of polynomial division.
The question is attached to the picture please refer to the picture.
We have a polynomial:
x³ - 3x² - 10x + 24
Which is divided by (x - 2)
We can divide using the synthetic division:
The first row: 1 -3 -10 24
The entry outside the row: 2
The missing value which is 2
k = 2
Thus, the value of k is 2 if the polynomial x³ - 3x² - 10x + 24 is divided by the expression (x - 2).
Learn more about the synthetic division here:
brainly.com/question/11850611
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A because your not dividing it by 5 your subtracting 5 from k divided by 2
X + y = 23000
x.08 + y.09 = 2040
x = 23000 - y
(23000 - y) .08 + y.09 = 2040
1840 - .08y + .09y = 2040
Simplify
.01y = 200
Simplify
Y = 20000
X + 20000 = 23000
X = 3000
CHECK: .08x + .09(20000) = 2040
.08x + 1800 = 2040
Simply .08x = 240
X = 3000
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.
