Answer:
Step-by-step explanation:
We are factoring
So:
((2•5^2x^2) + 485x) - 150
Pull like factors :
50x^2 + 485x - 150 = 5 • (10x^2 + 97x - 30)
Factor
10x^2 + 97x - 30
Step-1: Multiply the coefficient of the first term by the constant 10 • -30 = -300
Step-2: Find two factors of -300 whose sum equals the coefficient of the middle term, which is 97.
-300 + 1 = -299
-150 + 2 = -148
-100 + 3 = -97
-75 + 4 = -71
-60 + 5 = -55
-50 + 6 = -44
-30 + 10 = -20
-25 + 12 = -13
-20 + 15 = -5
-15 + 20 = 5
-12 + 25 = 13
-10 + 30 = 20
-6 + 50 = 44
-5 + 60 = 55
-4 + 75 = 71
-3 + 100 = 97
Step-3: Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 100
10x^2 - 3x + 100x - 30
Step-4: Add up the first 2 terms, pulling out like factors:
x • (10x-3)
Add up the last 2 terms, pulling out common factors:
10 • (10x-3)
Step-5: Add up the four terms of step 4:
(x+10) • (10x-3)
Which is the desired factorization
Thus your answer is
Answer:
Step-by-step explanation:
Eek! Let's give this a go. Things we know:
acceleration of Bond in free fall is -9.8 m/s/s
velocity of the truck is 25 m/s
displacement Bond will travel when he jumps is -10 m
What we are looking for is the time it will take him to hit the top of the truck, knowing that the truck can travel from one pole to the next in 1 second.
Our displacement equation is
Δx = v₀t + 1/2at²
Filling in we have
Simplifying we get
This is a quadratic that needs to be solved however you personally solve quadratics. When you do that, you find that the times it will take Bond to drop that displacement is either -.37 seconds or 5.47 seconds. Many things in physics can be negative, like velocity and acceleration, but time NEVER will be. So it takes Bond 5.5 seconds to drop to the roof of the moving truck. That means that he needs to jump when the truck is between the 5th and the 6th poles away from him.
Good luck with this!
Cheers!
Answer:
There are 15 letters, but if the two A's must always be together, that's the same as if they're just one letter, so our "base count" is 14! ; note that this way of counting means that we also don't need to worry about compensating for "double counting" identical permutations due to transposition of those A's, because we don't "count" both transpositions. However, that counting does "double count" equivalent permutations due to having two O's, two N's, and two T's, so we do need to compensate for that. Therefore the final answer is 14!/(23)=10,897,286,400
Add 70+93, then divide by the number of numbers, which is 2.
70+90=163
163/2=81.5
Hope this helps!
