i don't know specifically what you are looking for but:
Direction: Opens Up
Vertex: (1, -1)
Axis of Symmetry: x = 1
and i attached the graph
Answer: x = √(((-F - 4)^2)/16 - 1/2) + (F + 4)/4 or x = (F + 4)/4 - √(((-F - 4)^2)/16 - 1/2)
Step-by-step explanation:
Solve for x:
F x = 2 x^2 - 4 x + 1
Subtract 2 x^2 - 4 x + 1 from both sides:
-2 x^2 + F x + 4 x - 1 = 0
Collect in terms of x:
-1 + x (F + 4) - 2 x^2 = 0
Divide both sides by -2:
1/2 + 1/2 x (-F - 4) + x^2 = 0
Subtract 1/2 from both sides:
1/2 x (-F - 4) + x^2 = -1/2
Add 1/16 (-F - 4)^2 to both sides:
1/16 (-F - 4)^2 + 1/2 x (-F - 4) + x^2 = 1/16 (-F - 4)^2 - 1/2
Write the left hand side as a square:
(1/4 (-F - 4) + x)^2 = 1/16 (-F - 4)^2 - 1/2
Take the square root of both sides:
1/4 (-F - 4) + x = √(1/16 (-F - 4)^2 - 1/2) or 1/4 (-F - 4) + x = -√(1/16 (-F - 4)^2 - 1/2)
Subtract 1/4 (-F - 4) from both sides:
x = √(((-F - 4)^2)/16 - 1/2) + (F + 4)/4 or 1/4 (-F - 4) + x = -√(1/16 (-F - 4)^2 - 1/2)
Subtract 1/4 (-F - 4) from both sides:
Answer: x = √(((-F - 4)^2)/16 - 1/2) + (F + 4)/4 or x = (F + 4)/4 - √(((-F - 4)^2)/16 - 1/2)
Answer:
15 and 20
Step-by-step explanation:
When dividing 88 by 5 we wil have;
88/5
= 17 3/5
= 17 + 0.6
= 17.6
So we are to find the two numbers that 17.6 falls in between
From the given option 17.6 falls between 15 and 20. Hence the required numbers are 15 and 20
Answer:
Step-by-step explanation:
Unless I'm reading this incorrectly, he throws 155 6's.
There are 1000 throws altogether (according to the table)
So the experimental probability is 155/1000 = 0.155
The answer is B. It is a bit tricky to read.
Percent Error
Item
Books
Approximate
value
75
Exact value
95
Ratio
Error
-20
Absolute error
20
Percent error
(100)
.
元
(100)
Percent error = |actual error - expected error/ expected error| * 100%
(|75-95|)/95 = 0.21
0.21 * 100% = 21% error
Answer: 120[4(x^6 + x^3 + x^4 + x) +7(x^7 + x^4 + x^5 + x^2)]
Step-by-step explanation:
=24x(x^2 + 1)4(x^3 + 1)5 + 42x^2(x^2 + 1)5(x^3 + 1)4
Remove the brackets first
=[(24x^3 +24x)(4x^3 + 4)]5 + [(42x^4 +42x^2)(5x^3 + 5)4]
=[(96x^6 + 96x^3 +96x^4 + 96x)5] + [(210x^7 + 210x^4 + 210x^5 + 210x^2)4]
=(480x^6 + 480x^3 + 480x^4 + 480x) + (840x^7 + 840x^4 + 840x^5 + 840x^2)
Then the common:
=[480(x^6 + x^3 + x^4 + x) + 840(x^7 + x^4 + x^5 + x^2)]
=120[4(x^6 + x^3 + x^4 + x) +7(x^7 + x^4 + x^5 + x^2)]