85001 round to the nearest ten thousand place

rodikova [14]

P(x) = –x4 + x3 + 7x2 − x − 6

The graph will look like a bell, or a downward facing parabola, since the dominant term is -x^4

The graph has a y-intercept of -6, which can be considered the fixed cost (cost of zero trees cut) and moves upward from there, eventually reaching a maximum at some positive number of trees and dropping again. As x gets very large, profit becomes a very large negative number. Since we can't have negative trees cut, we're only interested in positive x values

The company breaks even at P(x) = 0
0 = –x4 + x3 + 7x2 − x − 6, x greater than zero
0 = (x+2)(x+1)(x-1)(x-3)
The company breaks even at 1 tree and 3 trees.

4 0

3 years ago

Solve the inequality: 6(2y - 4) + 2.6 > 14.7

aleksandr82 [10.1K]

Answer:

y > 3.0083

Step-by-step explanation:

6(2y-4) +2.6 >14.7

=6 ×2y +6(-4) +2.6 >14.7

=12y - 24 + 2.6> 14.7

And then collect like terms

12y > 14.7 +24 -2.6

12y > 38.7 - 2.6

12y > 36.1

12. 12

y > 3.0083

S.S[4,5,6,7......]

3 0

3 years ago

We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance

MAVERICK [17]

Answer:

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

$h(t) = -16\cdot t^{2}+256\cdot t+1680$ (1)

Where:

$h$ - Height above the ground, measured in feet.

$t$ - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

$h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680$

$h(3\,s) = 2304\,ft$

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at $t = 5\,s$ and $t = 7\,s$ :

$h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680$

$h(5\,s) = 2560\,s$

$h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680$

$h(7\,s) = 2688\,s$

We notice that the small object ascends in the given interval.

$\Delta h = h(7\,s)-h(5\,s)$

$\Delta h = 128\,ft$

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that $h = 2640\,ft$ , then (1) is reduced into this second-order polynomial:

$-16\cdot t^{2}+256\cdot t-960=0$ (2)

All roots of the resulting equation come from the Quadratic Formula:

$t_{1} = 10\,s$ and $t_{2}= 6\,s$

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that $h = 0\,ft$ , then (1) is reduced into this second-order polynomial:

$-16\cdot t^{2}+256\cdot t +1680 = 0$ (3)

All roots of the resulting equation come from the Quadratic Formula:

$t_{1} = 21\,s$ and $t_{2} = -5\,s$

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

8 0

3 years ago

Point Q is located at (-9, 10).

stich3 [128]

Answer:

(-5, -1)

Step-by-step explanation:

Moving towards the right makes a number more positive for x and moving down makes numbers more negative for y

7 0

3 years ago

Read 2 more answers

Find the nth term of this sequence -4 -3 -2 -1 0 ...

AnnyKZ [126]

The answer to your question is:

+4

-4, -3, -2, -1, 0, 1, 2, 3, 4

3 0

3 years ago