An easy way to solve this is to convert 1/8 into a percentage then see from that point.
To turn 1/8 into a percentage, you must first get the decimal form of the fraction. In this case, 1/8 as a decimal is 0.125.
Now that we have the decimal form, we can divide the decimal over 100 (common numerical number to use when dividing) to get the percentage, which should be 12.5%.
Now, our new question is 'what's bigger, 80% or 12.5%?'
The answer is 80%, since it is larger than 12.5%.
Answer:
Square
Step-by-step explanation:
the quadrilateral obtained by joining the mid points of adjacent sides of a square is a square
There’s no pictures what shape
Answer:
81.76in^2
Step-by-step explanation:
Given data
Lenght of book= 11.2 inches
WIdth of book= 7.3 inches
We know that Area= Length* Width
Area= 11.2*7.3
Area= 81.76 in^2
Hence the area of the front of the book is is 81.76in^2
Since ![[0,4]=[0,1]\cup(1,4]](https://tex.z-dn.net/?f=%5B0%2C4%5D%3D%5B0%2C1%5D%5Ccup%281%2C4%5D)
, we can rewrite the integral as
Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:
Both integrals are quite immediate: you only need to use the power rule
to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.
