Answer: Eric spends $41.44 on the contact paper.
Step-by-step explanation:
The dimensions of the box are 9 inches, 11 inches, 8 inches.
The surface area of the box will be given by:
![S.A.=2(lb+bh+hl)\\\Rightarrow\ S.A.=2(9(11)+11(8)+8(9))\\=2(99+88+72)\\=2\times259\\=518\ inch^2](https://tex.z-dn.net/?f=S.A.%3D2%28lb%2Bbh%2Bhl%29%5C%5C%5CRightarrow%5C%20S.A.%3D2%289%2811%29%2B11%288%29%2B8%289%29%29%5C%5C%3D2%2899%2B88%2B72%29%5C%5C%3D2%5Ctimes259%5C%5C%3D518%5C%20inch%5E2)
Since, the contact paper costs 8 cents per square inch.
The cost of covering the box=![8\times518=\4144\ cents](https://tex.z-dn.net/?f=8%5Ctimes518%3D%5C4144%5C%20cents)
Since, 1 $ = 100 cents or 1 cent =0.01 dollar.
Therefore, The cost of covering the box= ![=4144\times0.01=\$41.44](https://tex.z-dn.net/?f=%3D4144%5Ctimes0.01%3D%5C%2441.44)
Step-by-step explanation:
![\frac{140}{130} = 1.07692308 \approx \: 1.08 \\ \\ 104\% = \frac{104}{100} = 1.04 \\ \\ \because \: 1.08 > 1.04 \\ \\ \therefore \: \frac{140}{130} > 104\% \\](https://tex.z-dn.net/?f=%20%5Cfrac%7B140%7D%7B130%7D%20%20%3D%201.07692308%20%5Capprox%20%5C%3A%201.08%20%5C%5C%20%20%5C%5C%20%20104%5C%25%20%3D%20%5Cfrac%7B104%7D%7B100%7D%20%20%3D%201.04%20%5C%5C%20%20%5C%5C%20%20%5Cbecause%20%5C%3A%201.08%20%3E%201.04%20%5C%5C%20%20%20%5C%5C%20%20%5Ctherefore%20%5C%3A%20%20%5Cfrac%7B140%7D%7B130%7D%20%20%3E%20104%5C%25%20%20%5C%5C%20)
Thus, option A > is the correct answer.
A of rect: L*W= 5x3= 15
A of sq= s^2= 3^2=9
He is incorrect because 15 is not double of 9, if the area of the rect is really double the area of the square then its area should be 18 not 15.
Answer:
first option
Step-by-step explanation:
Always remember to graph a diagram :)
Answer:
$29364.95
Step-by-step explanation:
The formula to apply is;
![A=P(1+\frac{r}{n} )^{nt}](https://tex.z-dn.net/?f=A%3DP%281%2B%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bnt%7D)
where ;
A=amount at the end
P=principal amount
r=rate of interest as a decimal
n=number of compoundings a year
t=total number of years
Given in the question
A=$55000
t=8 years
n=2
r=0.08
P=?
Substitute the values in the equation;
![A=P(1+\frac{r}{n} )^{nt} \\\\\\55000=P(1+\frac{0.08}{2} )^{8*2} \\\\\\55000=P(1+0.04)^{16} \\\\\\55000=P(1.04)^{16} \\\\\\55000=1.8729P\\\\\\\frac{55000}{1.8729} =\frac{1.8729P}{1.8729} \\\\\\29364.95=P](https://tex.z-dn.net/?f=A%3DP%281%2B%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bnt%7D%20%5C%5C%5C%5C%5C%5C55000%3DP%281%2B%5Cfrac%7B0.08%7D%7B2%7D%20%29%5E%7B8%2A2%7D%20%5C%5C%5C%5C%5C%5C55000%3DP%281%2B0.04%29%5E%7B16%7D%20%5C%5C%5C%5C%5C%5C55000%3DP%281.04%29%5E%7B16%7D%20%5C%5C%5C%5C%5C%5C55000%3D1.8729P%5C%5C%5C%5C%5C%5C%5Cfrac%7B55000%7D%7B1.8729%7D%20%3D%5Cfrac%7B1.8729P%7D%7B1.8729%7D%20%5C%5C%5C%5C%5C%5C29364.95%3DP)
Checking the answer
![A=P(1+\frac{r}{n} )^{nt} \\\\\\A=29364.95(1+\frac{0.08}{2} )^{16} \\\\\\A=29364.95(1.04)^{16} \\\\\\A=55000](https://tex.z-dn.net/?f=A%3DP%281%2B%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bnt%7D%20%5C%5C%5C%5C%5C%5CA%3D29364.95%281%2B%5Cfrac%7B0.08%7D%7B2%7D%20%29%5E%7B16%7D%20%5C%5C%5C%5C%5C%5CA%3D29364.95%281.04%29%5E%7B16%7D%20%5C%5C%5C%5C%5C%5CA%3D55000)