HELP!!!!!Saquina dilates a digital photograph by a factor of 1.75 to create a new digital photograph similar to her original photograph.The perimeter of Saquina's new photograph is times larger than the perimeter of her original photograph, and the area of her new photograph is times larger than the area of her original photograph.

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   [tex]\bf 1.\underline{75}\implies \cfrac{175}{1\underline{00}}\implies \cfrac{7}{4}\implies \stackrel{ratio}{7:4}[/tex]

now, we know then the ratio dimensions for the new photograph, 

[tex]\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array} \\\\ -----------------------------\\\\[/tex]

[tex]\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}[/tex]