If you are asking for how many beads she used in total for the 12 necklaces it is
336. If you are asking how many beads she had before making the necklaces it is 471. If you are asking how many necklaces can she make with the leftover beads it is about 4. Hope this helps plz mark a crown.
Answer:
the solution is A
Step-by-step explanation:
![\sqrt[3]{27a^{3}b^{7} } \\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27a%5E%7B3%7Db%5E%7B7%7D%20%20%7D%20%5C%5C)
![\sqrt[3]{(3)^3(a^3)(b^6)b}\\\\3ab^2\sqrt[3]{b}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%283%29%5E3%28a%5E3%29%28b%5E6%29b%7D%5C%5C%5C%5C3ab%5E2%5Csqrt%5B3%5D%7Bb%7D)
Answer:
32, 36, 40
Step-by-step explanation:
8x + 9x + 10x = 108
27x = 108
x = 4
4 × 8 = 32 - 1st side
4 × 9 = 36 - 2nd side
4 × 10 = 40 - 3rd side
Answer:
Approximately 51 Yards
Step-by-step explanation:
You can solve this by cutting the square in half to make a right triangle. Than, you can use the Pythagorean theorem to solve for the hypotenuse.
a^2+b^2=c^2
36^2+36^2=c^2
1296+1296=c^2
2592=c^2
50.9116
Round to the nearest whole number since the question says approximately: 51
<h3>
Answer: 2p + 3q</h3>
Work Shown:
log(200) = log(2^3*5^2)
log(200) = log(2^3) + log(5^2)
log(200) = 3*log(2) + 2*log(5)
log(200) = 3*q + 2*p
log(200) = 2p + 3q
The log rules I used were
log(A*B) = log(A)+log(B)
log(A^B) = B*log(A)
