Answer: 99, 100, 101. 102, 103
Step-by-step explanation:
Divide 505÷5 to get the middle integer, 101.
From that, subtract 2 to get the lowest integer, 99.
Add 101 + 2 for the greatest integer, 103. Fill in the other two integers and put them in consecutive order.
The answer is B, when you add ∠MQR and ∠LQR you will get <span>180°.</span>
Answer:
x>-5
Step-by-step explanation:
-3 from both sides
Answer:
So possibilities of zeroes are:
Positive Negative Imaginary
1 1 2
3 1 0
Zeroes = -1.4549, 1.2658, 0.34457-1.0503i, 0.34457+1.0503i.
Step-by-step explanation:
Note: Descartes' Rule of Signs is used to find the signs of zeroes not the exact value.
The given function is
Degree of polynomial is 4 so number of zeroes is 4.
There are three sign changes, so there are either 3 positive zeros or 1 positive zero.
Now, put x=-x in f(x).
There is one variation in sign change, so there is 1 negative zero.
So possibilities of zeroes are:
Positive Negative Imaginary
1 1 2
3 1 0
Using graphing calculator the zeroes of given function are -1.4549, 1.2658, 0.34457-1.0503i and 0.34457+1.0503i.
You simply cross multiply. The correct answer is B. I know this becase I had the same exact question in my Staar review packet.