Answer:
C
Step-by-step explanation:
The data is skewed, and the lowest number of crackers in a package was 7
Answer:
-1, 1
13, 15
Step-by-step explanation:
x and x+2 are the integers
- x*(x+2)= 7(x+x+2) -1
- x²+2x= 14x+14-1
- x² - 12x -13= 0
Roots of the quadratic equation are: -1 and 13.
So the integers pairs are: -1, 1 and 13, 15
So to solve problems like these you can work out each equation by substituting the x and y with the coords from the vertex (2,-4) and which ever one is true is the corresponding equation.
Lets try the first one
A) y = 2( x-2)^2-4 this would become
-4 = 2(2-2)^2-4 we solve this and get
-4 = 2(0)-4
-4 = -4
so it seems like A is the correct answer, of course we'd wanna check out the other answers just to be sure.
Lets try one and do C
C) -2 = 2(-4-2)^2+2
-2 = 2(-8)^2+2
-2 = 130
These aren't equal so this can't be our equation
So you can also do one more just to be super sure you have the right answer but I think A is the correct one :)
Answer:
they mean you repeat that behavior
Step-by-step explanation:
you constantly are on your phone
Answer:
20. AB = 42
21. BC = 28
22. AC = 70
23. BC = 20.4
24. FH = 48
25. DE = 10, EF = 10, DF = 20
Step-by-step explanation:
✍️Given:
AB = 2x + 7
BC = 28
AC = 4x,
20. Assuming B is between A and C, thus:
AB + BC = AC (Segment Addition Postulate)
2x + 7 + 28 = 4x (substitution)
Collect like terms
2x + 35 = 4x
35 = 4x - 2x
35 = 2x
Divide both side by 2
17.5 = x
AB = 2x + 7
Plug in the value of x
AB = 2(17.5) + 7 = 42
21. BC = 28 (given)
22. AC = 4x
Plug in the value of x
AC = 4(17.5) = 70
✍️Given:
AC = 35 and AB = 14.6.
Assuming B is between A and C, thus:
23. AB + BC = AC (Segment Addition Postulate)
14.6 + BC = 35 (Substitution)
Subtract 14.6 from each side
BC = 35 - 14.6
BC = 20.4
24. FH = 7x + 6
FG = 4x
GH = 24
FG + GH = FH (Segment Addition Postulate)
(substitution)
Collect like terms


Divide both sides by -3

FH = 7x + 6
Plug in the value of x
FH = 7(6) + 6 = 48
25. DE = 5x, EF = 3x + 4
Given that E bisects DF, therefore,
DE = EF
5x = 3x + 4 (substitution)
Subtract 3x from each side
5x - 3x = 4
2x = 4
Divide both sides by 2
x = 2
DE = 5x
Plug in the value of x
DE = 5(2) = 10
EF = 3x + 4
Plug in the value of x
EF = 3(2) + 4 = 10
DF = DE + EF
DE = 10 + 10 (substitution)
DE = 20