Since the square roots are both six and cannot be simplified any more you just treat it like a normal subtraction problem
so -5 - 2= -7
so it would be -7√2
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
50÷(-25)x(4)
50/-25= -2
-2 x 4= -8
Sorry I realized I put 24 Oops...
But Ctapia037 is right.
