let's first off convert those mixed fractions to improper fractions, then get their difference.
![\bf \stackrel{mixed}{1\frac{1}{2}}\implies \cfrac{1\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{3}{2}}~\hfill \stackrel{mixed}{2\frac{1}{10}}\implies \cfrac{2\cdot 10+1}{10}\implies \stackrel{improper}{\cfrac{21}{10}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{21}{10}-\cfrac{3}{2}\implies \stackrel{\textit{using the LCD of 10}}{\cfrac{(1)21-(5)3}{10}}\implies \cfrac{21-15}{10}\implies \cfrac{6}{10}\implies \cfrac{3}{5}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B1%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B1%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B3%7D%7B2%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B1%7D%7B10%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%2010%2B1%7D%7B10%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B21%7D%7B10%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Ccfrac%7B21%7D%7B10%7D-%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Busing%20the%20LCD%20of%2010%7D%7D%7B%5Ccfrac%7B%281%2921-%285%293%7D%7B10%7D%7D%5Cimplies%20%5Ccfrac%7B21-15%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B6%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B3%7D%7B5%7D)
now, the original amount, 3/2, if that is the 100%, what is 3/5 off of it in percentage?
Answer:
1865.16 m^2
Step-by-step explanation:
first let 's find radius of a circle
radius=diameter/2
=18/2
=9 m
area of a cylinder=2πrh+2πr2
=2*3.14*9*24 + 2*3.14*(9)^2
=1356.48 + 2*3.14*81
=1356.48 + 508.68
=1865.16 m^2
Square and Rhombus are the following quadrilaterals have diagonals that are always perpendicular to each other.
C. Square
D. Rhombus
<u>Step-by-step explanation:</u>
This implies the diagonals of a square and rhombus are perpendicular. The diagonals of a square and rhombus are a similar length. In elementary geometry, the property of being opposite is the connection between two lines which meet at a right angle. The property stretches out to other related geometric items.
Principally Perpendicular lines will be lines that cross at a right (90 degrees) edge. so when it goes under shape rhombus and square have the equivalent of a considerable number of sides parallelly.
Answer:
1. HL
2. SAS
3. SSS
Step-by-step explanation:
Triangles ABR and ACR share side AR (hypotenuse of two right triiangles).
Angles ABR and ACR are right angles.
Sides AB and AC are congruent.
Sides BR and CR are congruent.
1. You can use HL theorem, because two triangles have congruent pair of legs and congruent hypotenuses.
2. You can use SAS theorem, because two triangles have two pairs of congruent legs and a pair of included right angles between these legs.
3. You can use SSS theorem, because two triangles have two pairs of congruent legs and congruent hypotenuses.
Answer:
D is the answer just took the test!!!!
Step-by-step explanation:
