Answer:
37 dimes and 10 nickels
Step-by-step explanation:
let d = # dimes
let n = # nickels
we can set up a system of equations:
n + d = 47
.05n + .10d = 4.2
if we solve the first equation for 'n' we get:
n = 47-d
now we can substitute this in for 'n' in the second equation:
.05(47-d) + .10d = 4.2
2.35 - .05d + .10d = 4.2
2.35 + .05d = 4.2
subtract 2.35 from each side to get:
.05d = 1.85
d = 1.85÷.05
d = 37
if d+n = 47 and d=37 then n = 10
Check:
.05(10) + .1(37) should equal 4.2
.50 + 3.7 = 4.2 [It Checks Out]
20 comic books is the right awnser
The area <em>A</em> of a trapezoid with height <em>h</em> and bases <em>b</em>₁ and <em>b</em>₂ is equal to the average of the bases times the height:
<em>A</em> = (<em>b</em>₁ + <em>b</em>₂) <em>h</em> / 2
We're given <em>A</em> = 864, <em>h</em> = 24, and one of the bases has length 30, so
864 = (<em>b</em>₁ + 30) 24 / 2
864 = (<em>b</em>₁ + 30) 12
864 = (<em>b</em>₁ + 30) 12
72 = <em>b</em>₁ + 30
<em>b</em>₁ = 42
Answer:
Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, ![[ab]_K](https://tex.z-dn.net/?f=%20%5Bab%5D_K%20)
is equal to ![[ba]_K](https://tex.z-dn.net/?f=%20%5Bba%5D_K%20)
, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.
Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, ![[ab]_H = [ba]_H](https://tex.z-dn.net/?f=%20%5Bab%5D_H%20%3D%20%5Bba%5D_H%20)
. Since a and b were generic elements of H, then H/G is abelian.
Answer:100
Step-by-step explanation:
