Answer:
in △abc, the sides ab and ac are produced to ∠p and ∠q respectively. if the bisectors of ∠pbc and ∠qcb intersect at a point o. prove that ∠boc = 90 - ½ ∠a.
Answer:
first one is 3 and second one is idk
Step-by-step explanation:
General Idea:
Integers in real life are used to denote both positive and negative numbers. For example to represent profit, gain, increase, deposit, Income we use positive integers and to represent loss, expenditure, decrease, withdrawal we use negative integers.
Conclusion:
The integer describes gaining 5 pounds is +5.
Answer:
The solution is:
Part A.
which is sqrt(5)^7k/3[/tex]
Part B. k = 18/7
Step-by-step explanation:
Part A.
To solve this part, we're going two use THREE important properties of exponents:
1. ![(x^{n})^{m} = x^{nm}](https://tex.z-dn.net/?f=%28x%5E%7Bn%7D%29%5E%7Bm%7D%20%3D%20x%5E%7Bnm%7D)
2. ![\frac{x^{n}}{x^{m}} = x^{n-m}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7Bn%7D%7D%7Bx%5E%7Bm%7D%7D%20%3D%20x%5E%7Bn-m%7D)
3. ![\sqrt[n]{x^{m}} = x^{\frac{m}{n} }](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5E%7Bm%7D%7D%20%3D%20x%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%20%7D)
Let's work the numerator using the properties 1, 2 and 3:
![(\sqrt{5}^{3} )^{\frac{k}{9} } } = (\sqrt{5}^{3\frac{k}{9}}) = (\sqrt{5}^{\frac{k}{3}})](https://tex.z-dn.net/?f=%28%5Csqrt%7B5%7D%5E%7B3%7D%20%29%5E%7B%5Cfrac%7Bk%7D%7B9%7D%20%7D%20%7D%20%20%3D%20%28%5Csqrt%7B5%7D%5E%7B3%5Cfrac%7Bk%7D%7B9%7D%7D%29%20%3D%20%28%5Csqrt%7B5%7D%5E%7B%5Cfrac%7Bk%7D%7B3%7D%7D%29)
Let's work the denominator using the properties 1, 2 and 3:
![(\sqrt{5}^{6} )^{-\frac{k}{3} } } = (\sqrt{5}^{6\frac{k}{3}}) = (\sqrt{5}^(2k))](https://tex.z-dn.net/?f=%28%5Csqrt%7B5%7D%5E%7B6%7D%20%29%5E%7B-%5Cfrac%7Bk%7D%7B3%7D%20%7D%20%7D%20%20%3D%20%28%5Csqrt%7B5%7D%5E%7B6%5Cfrac%7Bk%7D%7B3%7D%7D%29%20%3D%20%28%5Csqrt%7B5%7D%5E%282k%29%29)
Now dividing the numerator by the denominator:
![\sqrt{5}^{\frac{k}{3}-(-2k)})=\sqrt{5}^{\frac{7k}{3}})](https://tex.z-dn.net/?f=%5Csqrt%7B5%7D%5E%7B%5Cfrac%7Bk%7D%7B3%7D-%28-2k%29%7D%29%3D%5Csqrt%7B5%7D%5E%7B%5Cfrac%7B7k%7D%7B3%7D%7D%29)
Part B
if ![5^{\frac{3}{2} } 5^{\frac{3}{2}} = \sqrt{5}^{\frac{7k}{3}})](https://tex.z-dn.net/?f=5%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%205%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%20%3D%20%5Csqrt%7B5%7D%5E%7B%5Cfrac%7B7k%7D%7B3%7D%7D%29)
Then:
![5^{3} = 5^{\frac{7k}{6}})](https://tex.z-dn.net/?f=5%5E%7B3%7D%20%3D%205%5E%7B%5Cfrac%7B7k%7D%7B6%7D%7D%29)
So ![\frac{7k}{6}} = 3](https://tex.z-dn.net/?f=%5Cfrac%7B7k%7D%7B6%7D%7D%20%3D%203)
Solving for k, we have:
k = 18/7
Add all the numbers in the list then divide it by how many numbers are in that list. Ex. 5,8,1,4 Would be: 5+8+1+4= 18 divided by 4 which is 4.5