What is the value of a cube shaped box with edges 6 cm. in length?
2 answers:
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Answer:2.634 in./s
Step-by-step explanation:
Given
Length (L)=18 in.
width (B)=12 in.
height(h)=4 in.
Volume of water is increasing at a rate of 569
Volume of a box is given by=LBh
and
as L&B are constant therefore
Answer:
D. 1 real; 4 complex
Step-by-step explanation:
1. Calculate the total number of roots
The fundamental theorem of algebra states that any polynomial of degree n has n roots.
Your polynomial is fifth degree, so it has five roots.
2. Calculate the number of real roots
We can use Descartes' rule of signs to determine the number of real roots:
The number of positive real roots is the same as the number of changes in the sign of the coefficients of ƒ(x) or less than by an even number.
The number of negative real roots is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number.
(a) Number of positive real roots
The coefficients of ƒ(x) are
+1 +3 +8 +14 +16 + 8
There are no changes of sign, so there are no positive real roots.
(b) Number of negative real roots
(i) Find f(-x)
f(-x) = -x⁵+ 3x⁴ - 8x³ + 14x² - 16x + 8
(ii) Count the changes of sign
The coefficients of f(-x) are
-1 ∥ +3 ∥ -8 ∥+14 ∥ -16 ∥ +8
There are five changes of sign.
Thus, the number of negative roots is five, three, or one.
So far, we know that we have five roots. None is positive, and there could be one, three, or five negative real roots.
3. Confirm by sketching the graph
The real roots are the points at which the graph crosses or touches the x-axis.
The graph (see below) crosses the x-axis at only one point.
Thus. we have
- one real root (negative)
- four complex roots
