22000=15000(1+.089/4)^4t
22/15=(1.02225)^4t
log(22/15)=4tlog1.02225
(log(22/15)/4log1.02225)=t
(.16633/.038228)=t
t=4.35 years or 4 years and approximately 4 months
The prime factors of 6 are 2 and 3. So for a number to be divisible by 6,it must also be divisible by 2 and 3.Therefore,we need to check if a number is even and then check if the sum of the digits is divisible by 3
Answer:
AAS
Step-by-step explanation:
I believe
Answer:
![\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%7D)
Step-by-step explanation:
Our expression is:
.
Let's focus on the cube root of 81 first. What's the prime factorisation of 81? It's simply: 3 * 3 * 3 * 3, or
. Put this in for 81:
![\sqrt[3]{81} =\sqrt[3]{3^3*3}=\sqrt[3]{3^3} *\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B81%7D%20%3D%5Csqrt%5B3%5D%7B3%5E3%2A3%7D%3D%5Csqrt%5B3%5D%7B3%5E3%7D%20%2A%5Csqrt%5B3%5D%7B3%7D)
We know that the cube root of 3 cubed will cancel out to become 3, but the cube root of 3 cannot be further simplified, so we keep that. Our outcome is then:
![\sqrt[3]{3^3} *\sqrt[3]{3}=3\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3%5E3%7D%20%2A%5Csqrt%5B3%5D%7B3%7D%3D3%5Csqrt%5B3%5D%7B3%7D)
Now, let's multiply this by 1/3, as shown in the original problem:
![\frac{1}{3}* 3\sqrt[3]{3}=\sqrt[3]{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%2A%203%5Csqrt%5B3%5D%7B3%7D%3D%5Csqrt%5B3%5D%7B3%7D)
Thus, the answer is
.
<em>~ an aesthetics lover</em>
440 yards in 1/4 of a mile