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
Answer: C - Only the fifth-grade data points are distributed fairly evenly.
Step-by-step explanation: Just took the test and that is the right anwser.
Answer:
x = 17, MN = 11
Step-by-step explanation:
Given 2 secants from an external point to a circle, then
The product of the external part and the whole of one secant is equal to the product of the external part and the whole of the other secant.
(5)
7(7 + x) = 8(8 + 13) = 8 × 21 = 168 ( divide both sides by 7 )
7 + x = 24 ( subtract 7 from both sides )
x = 17
(6)
9(9 + 2x - 7) = 10(10 + 8)
9(2x + 2) = 10 × 18 = 180 ( divide both sides by 9 )
2x + 2 = 20 ( subtract 2 from both sides )
2x = 18 ( divide both sides by 2 )
x = 9
Then
MN = 2x - 7 = 2(9) - 7 = 18 - 7 = 11
As the width gets larger so does the total area
