The true statement will be A.
Answer:
*rolls eyes* I didn't think this would be the community where moderators are missed. Oh well, she clearly doesn't need help with differential calculus or Python so I wish her best.
Step-by-step explanation:
Answer:
a = 1, b = 6
Step-by-step explanation:
The equation given is as follows;
P(x) = a·x³ + b·x² + 3·x - 10
The above equation has a factor of x + 5, therefore, we have;
P(-5) = 0 = a·(-5)³ + b·(-5)² + 3·(-5) - 10
-125·a + 25·b + (-15) - 10 = 0
-125·a + 25·b - 25 = 0
-125·a + 25·b = 25...........(1)
Also, we are given that;
a·x³ + b·x² + 3·x - 10 divided by x + 1 as a remainder, R = -8, therefore;
P(-1) = -8 = a·(-1)³ + b·(-1)² + 3·(-1) - 10
-a + b - 13 = -8
-a + b = -8 + 13 = 5
-a + b = 5............................(2)
Multiply equation (2) by 25 and subtract from (1) gives
-125·a + 25·b - 25(-a + b) = 25 - 25×5
-100·a = 25 - 125 = -100
a = 1
Therefore, from equation (2) we have;
-1 + b = 5
b = 5 + 1 = 6.
Answer:
C, E, and F
Explanation:
There are two ways to answer this question. First, you could simply input each answer into both equations to see which one works but that would take quite a long time.
A better way is to simply solve each equation for x.
You could rewrite
2x + 7 < -3
as
2x + 7 = -3
and solve:
Subtract 7 from both sides
Now divide both sides by 2
Now we can simply replace the equals sign with the inequality
x < -5
Where you can run into trouble is if you have to multiply or divide by a negative number across the equation, you must flip the inequality sign. It's best to leave it there to remind you, but I switched it out just to show that it's no different than a typical algebraic equation.
Now, we know that x can be any value less than -5. Let's find out the same thing for the second equation:
Now we know that x must be less than 1 for the second equation. So, now we can choose the answers that are both less than -5 and less than 1.
These answers are:
C. -10
E. -8.24
and
F. -15/2 which is -7.5
NOTE: -5 is equal to but not less than -5 so G is not included.