P(x) = (x + 3)Q(x) + R
<span> P(-3) = (0)Q(x) + R </span>
<span>i P(-3) = R </span>
<span>Now, P(-3) = (-3)⁴ - 9(-3)³ - 5(-3)² - 3(-3) + 4
</span>So
<span>P(-3) = 81 + 243 - 45 + 9 + 4 </span>
<span>=. 292
hope it helps</span>
Answer:
6. slope= -4/5 Y-intercept is 3
7.
a.y=-2x
b.y=1x+1
c.5/2x-1
d.y=4x+3
e.y=-3/2x-5
Step-by-step explanation:
Answer:
After the melting there were one and three fifths of snow left.
Step-by-step explanation:
Given that a snowstorm dropped three and two-fifths inches of snow, and the next day one and four-fifths inches melted, the following calculation must be performed to determine how much snow was left:
2/5 = 0.40
4/5 = 0.80
3.40 - 1.80 = X
1.60 = X
0.60 = 3/5
Thus, after the melting there were one and three fifths of snow left.
Answer:
-x^2 - 20x -4 = 0
Step-by-step explanation:
I assume you meant (x-8)(2x+3) = (3x-5)(x+4).
Perform the two indicated multiplications and then combine like terms:
2x^2 + 3x - 16x - 24 = 3x^2 + 12x - 5x - 20
Combine the x terms on each side:
2x^2 - 13x - 24 = 3x^2 + 7x - 20
Subtract 3x^2 from both sides:
-x^2 - 13x - 24 = 7x - 20
Subtract 7x from both sides:
-x^2 - 20x - 24 = -20
Add 20 to both sides:
-x^2 - 20x -4 = 0 This is the desired quadratic equation.
Answer:
Step-by-step explanation:
The altitude of each plane can be expressed by an equation that adds the initial altitude and the product of time and its rate of climb.
plane A: y = 4057 +60.75t
plane B: y = 5000 +40.25t
__
We want to find the time when their altitudes are the same:
4057 +60.75t = 5000 +40.25t . . . . equate the expressions for y
20.50t = 943 . . . . . . . . . subtract (40.24t +4057)
t = 46 . . . . . . . . . . . . divide by the coefficient of t
At t=46, the altitude of the planes will be ...
y = 5000 +40.25(46) = 5000 +1851.5 = 6851.5
The planes are at the same altitude after 46 seconds.
Both planes will be at an altitude of 6851.5 feet.
