Answer:
assuming its an annual interest
Okay so 6 percent interest, the bank is paying you.
So with this it’s 6 percent of 1500 and add it to 1500.
You can always find 6 percent of 1500 and then add but here’s a short cut.
Your principle (beginning) balance is 1500.
That’s already 100 percent since thats yoru original value.
You then get added 6 percent interest.
We are jsut adding 6 percent to 100 percent so 106 percent.
Now we solve normally and you’d get the answer faster.
106 percent is 106/100 or 1 3/5 or 1.06
now we multiply
1500 * 1.06 = 1590
Your final balance would be 1590 after the 6 percent interest is added.
She 56/92 6-5=1 and 567-1=566
Answer:
9, if you're asking for only the number of it. But If it needs more then please elaborate.
Step-by-step explanation:
4×9+7×9=(4+7)×9
36+63=11*9
99=99
Order Of Operation Helps!
Answer:
Option d) 5 to the power of negative 5 over 6 is correct.
![\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B%5Cbf%205%7D%20%5Ctimes%20%5Csqrt%7B%5Cbf%205%7D%7D%7B%5Csqrt%5B3%5D%7B%5Cbf%205%5E%7B%5Cbf%205%7D%7D%7D%3D%205%5E%7B%5Cfrac%7B%5Cbf%20-5%7D%7B%5Cbf%206%7D%7D)
Above equation can be written as 5 to the power of negative 5 over 6.
ie, 
Step-by-step explanation:
Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.
It can be written as below
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%20%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%20%5Cdfrac%7B5%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5Ctimes%205%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B5%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%20%5Cdfrac%7B5%5E%7B%5Cfrac%7B1%7D%7B3%7D%2B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B5%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%20%5Cdfrac%7B5%5E%7B%5Cfrac%7B2%2B3%7D%7B6%7D%7D%7D%7B5%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%205%5E%7B%5Cfrac%7B5%7D%7B6%7D%7D%20%5Ctimes%205%5E%7B%5Cfrac%7B-5%7D%7B3%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%205%5E%7B%5Cfrac%7B5-10%7D%7B6%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B5%5E5%7D%3D%205%5E%7B%5Cfrac%7B-5%7D%7B6%7D%7D)
Above equation can be written as 5 to the power of negative 5 over 6.
Answer:
Part A: 0
Part B: 0 or 1
Step-by-step explanation:
Part A: anything to the 0 power is 1 (x^0= x^y-y) if you solve 7^2 you get 49. 49^0= 1
Part B: Because it is already being raised to the power of 0, if x=0 it will remain equal to one (as explained above). Anything raised to the 1 power is itself, so 7^0= 1 and 1^1 =1
Hope this helped!
