Answer:
The function which models the scenario is
Initial thickness = 2 mm
Value of thickness per fold = 2(previous thickness)
A table can be created thus :
Number of folds(x) __ Thickness
0 _______________ 2
1 _______________ 2(2) = 4
2 _______________ 2(4) = 8
3 _______________ 2(8) = 16
4 ______________ 2(16) = 32
5 ______________ 2(32) = 64
The relation is an exponential function which can be modeled thus :
A = Initial thickness
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Answer:
A minimum of 10 dimes and 11 quarters is what Alexandra will have
Step-by-step explanation:
Let
d = number of dimes
q = number of quarters
Since she has 21 coins altogether,
d + q = 21------------------------equation 1
- If these coins are worth $3.75 then
0.10 x d + 0.25 x q = 3.75
- which is 0.10d +.25q =3.75
--------------------------equation 2
where $.10 is the value of one dime and $.25 is the value of one quarter
make d the subject of formula from equation 1 d = 21 -q----------equation 3
insert it in equation 2
0.10d +0.25q =3.75
0.10(21-q) + 0.25q = 3.75
0.1(21)-0.1q+0.25q=3.75
2.1 +0.15q = 3.75
0.15q = 3.75-2.1 = 1.65
q = 1.65/0.15 =165/15 =11
- since we have the value of q insert in equation 3
d = 21 - q
d = 21-11
d = 10
Alexandra has 10 dimes and 11 quarters.
from my calculation i can see that the a minimum of 10 dimes and 11 quarters is what Alexandra will have
X+2y=2 …(1)
x-2y=-2…(2)
(1)-(2): (x+2y)-(x-2y)=2- (-2)
4y=4
y=1
subs y=1 into(1)
x+2(1) =2
x=0
so x =0 y=1
the answer is C
Answer: 15 should be the answer hope this helped
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Step-by-step explanation:
Answer:
Can you explain more?
Step-by-step explanation:
