How many solutions does the equation have? 5(w−1)−2=5w+7A.0 B.1 C.infinitely many
2 answers:
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Answer:
E is not a subspace of
Step-by-step explanation:
E is not a subspace of
In order to see this, we must find two points (a,b), (c,d) in E such that (a,b) + (c,d) is not in E.
Consider
(a,b) = (1,1)
(c,d) = (-1,-1)
It is easy to see that both (a,b) and (c,d) are in E since 1*1>0 and (1-)*(-1)>0.
But (a,b) + (c,d) = (1-1, 1-1) = (0,0)
and (0,0) is not in E.
By the way, it can be proved that in any vector space all sub spaces must have the vector zero.
First off, let's convert the decimal to a fraction, notice, we have two decimals, so we'll use in the denominator, a 1 with two zeros then, two decimals, two zeros, thus 
now, we know then the ratio dimensions for the new photograph,
![\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\ \cfrac{7}{4}\implies \cfrac{4+3}{4}\implies \cfrac{4}{4}+\cfrac{3}{4}\implies 1+\boxed{\cfrac{3}{4}}\impliedby \textit{perimeter is }\frac{3}{4}\textit{ larger} \\\\\\ \stackrel{areas'~ratio}{\cfrac{s^2}{s^2}}\implies \cfrac{3^2}{4^2}\implies \cfrac{9}{16}\impliedby \textit{area is }\frac{9}{16}\textit{ larger than original}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%5Cqquad%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7Bs%5E2%7D%7D%7B%5Csqrt%7Bs%5E2%7D%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B7%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B4%2B3%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B4%7D%2B%5Ccfrac%7B3%7D%7B4%7D%5Cimplies%201%2B%5Cboxed%7B%5Ccfrac%7B3%7D%7B4%7D%7D%5Cimpliedby%20%5Ctextit%7Bperimeter%20is%20%7D%5Cfrac%7B3%7D%7B4%7D%5Ctextit%7B%20larger%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bareas%27~ratio%7D%7B%5Ccfrac%7Bs%5E2%7D%7Bs%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B3%5E2%7D%7B4%5E2%7D%5Cimplies%20%5Ccfrac%7B9%7D%7B16%7D%5Cimpliedby%20%5Ctextit%7Barea%20is%20%7D%5Cfrac%7B9%7D%7B16%7D%5Ctextit%7B%20larger%20than%20original%7D)
now, we know then the ratio dimensions for the new photograph,