Aquantity with an initial value of 6300 grows continuously at a rate of 60% per decade. what is the value of the quantity after 10 years, to the nearest hundredth?

A decade= 10 years

The problem says that the quantity "grows". This means the quantity is increasing. Therefore, we will be multiplying by a number over 1 because we are increasing the quantity past 100% which is = to 1. We are only increasing by 60% because in 10 years only one decade passes. 

60% +100%= 160%= 1.6

6300*1.6= 10080.00

We added two zeroes after the decimal place because the problem was asking us to round to the nearest hundredth.

10080.00

Comment below with any further questions you have!

The answer would be -3.9> -4.5

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)