If a recipe calls for 1.25 cups of sugar, what degree of accuracy would be sufficient for the markings on Jane's 2 cup measuring device she is using?
(1 point)
O iofacup
O½ofaaup
O of acup
O yofacup
The degree of accuracy that would be sufficient for the markings on Jane's 2 cup measuring device would be ½ of a cup.
The weight of a 20 dollar bill is 1.0 g. If Denny is using a count-by-weight machine to count how many 20 dollar bills are in a stack, which accuracy is most reasonable for the count-by-weight machine? (1 point)
0 5 g
O 1.0 g
00019 Dog
• 0.1 g
The accuracy that is most reasonable for the count-by-weight machine used by Denny would be 0.1 g.
Solve the equation below for variable v
A = + + rUf
(1 point)
I'm sorry, but the equation you provided is unclear. Could you please clarify the equation and provide any missing information or variables?
Solve the equation below for variable v
A = r+rvf
(1 point)
To solve the equation A = r + rvf for the variable v, we can first factor out v on the right side of the equation:
A = r(1 + vf)
Next, divide both sides of the equation by (1 + vf) to isolate v:
A/(1 + vf) = r
Finally, multiply both sides of the equation by (1 + vf):
v(A/(1 + vf)) = r(1 + vf)
Simplifying the right side gives:
vA/(1 + vf) = r + rvf
Distribute vA to get:
vA = r + rvf + rv^2f
Combine like terms on the right side:
vA = r(1 + vf + v^2f)
Divide both sides of the equation by r(1 + vf + v^2f):
vA/(r(1 + vf + v^2f)) = 1
Thus, the solution for v is:
v = 1/(r(1 + vf + v^2f))
Using the equation C = 27r if C=18.84 and m= 3.14.
Then r=
4
1
2
3
To solve the equation C = 27r, we can substitute the given values of C and solve for r.
Given:
C = 18.84
r = ?
Substituting C = 18.84 into the equation C = 27r:
18.84 = 27r
Now, solve for r:
Divide both sides of the equation by 27:
18.84/27 = r
0.6989 = r
Therefore, r ≈ 0.6989.
None of the answer choices provided (4, 1, 2, 3) are accurate.
Given d = 60 miles and v = 15 miles per hour
solve d - ut in terms of t then substitute values.
(1 point)
C t= 4 hours
( t= 15 hours
C 1= 60 hours
0 t= 900 hours