keeping in mind that when the logarithm base is omitted, the base 10 is assumed.
![\textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b \\\\[-0.35em] ~\dotfill\\\\ \log(x)=2\implies \log_{10}(x)=2\implies 10^2=x\implies 100=x](https://tex.z-dn.net/?f=%5Ctextit%7Bexponential%20form%20of%20a%20logarithm%7D%20%5C%5C%5C%5C%20%5Clog_a%28b%29%3Dy%20%5Cqquad%20%5Cimplies%20%5Cqquad%20a%5Ey%3D%20b%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Clog%28x%29%3D2%5Cimplies%20%5Clog_%7B10%7D%28x%29%3D2%5Cimplies%2010%5E2%3Dx%5Cimplies%20100%3Dx)
Answer:
If every line parallel to two lines intersects both regions in line segments of equal length, then the two regions have equal areas. In the case of your problem, every line parallel to the bases of the two parallelograms will intersect them in lines segments, each with a width of ℓ.
Lo siento, no hablo español
Answer:
8 cm
Step-by-step explanation:
... BO/OD = AO/OC = 3/1
Written another way, this is ...
... OD : BO = 1 : 3
Now, BD = OD + BO, so we have
... BD : BO = (OD +BO) : BO = (1 +3) : 3 = 4 : 3
That is, BD = 4/3 × BO
... BD = 4/3 × 6 cm = 8 cm
Answer:
1.81 inches
Step-by-step explanation:
Given that,
The circumference of the largest orange in a bag is 
and that of smallest orange is 
.
Difference = largest orange - smallest orange
= 
So, difference of the circumferences of the smallest orange and the largest orange is 1.81 inches
