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Blababa [14]
3 years ago
11

The sum of twice a number and 31

Mathematics
1 answer:
Andreyy893 years ago
3 0

Answer:

Let the unknown number be x ;

2(x) +31\\=2x+31

Step-by-step explanation:

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You are graphing Square ABCDABCDA, B, C, D in the coordinate plane. The following are three of the vertices of the square: A(4,
Kitty [74]

Answer:

D(4,-3)

Step-by-step explanation:

Given three of the vertices of the square: A(4, -7), B(8, -7),C(8, -3)

Let the coordinate of the fourth vertex be D(x,y).

We know that diagonals of a square are perpendicular bisector. So, the midpoint of both diagonals is the same.

The diagonals are BD and AC

Midpoint of BD = Midpoint of AC

\left(\dfrac{8+x}{2},\dfrac{-7+y}{2}\right) =\left(\dfrac{4+8}{2},\dfrac{-7+(-3)}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(\dfrac{12}{2},\dfrac{-10}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(6,-5\right)\\$Therefore$:\\\dfrac{8+x}{2}=6\\8+x=12\\x=12-8\\x=4\\$Similarly$\\\dfrac{y-7}{2}=-5\\y-7=-5*2\\y-7=-10\\y=-10+7=-3

The coordinates of the fourth vertex is D(4,-3)

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Susie is paid an annual salary of $71,500. Find Susie’s biweekly salary
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How do I write a comment on the data following the completion of the scatterplot
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Look at this cylinder:
Sedbober [7]
  • Height=h=8cm
  • Radius=r=4cm

We know

\boxed{\sf \star TSA_{(Cylinder)}=2\pi r(h+r)}

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=2\times \dfrac{22}{7}\times 4(8+4)

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{176}{7}(12)

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{2112}{7}

\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=301.7cm^2

Now

  • New Radius=2(4)=8cm
  • New Height=2(8)=16cm

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=2\times \dfrac{22}{7}\times 8(16+8)

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{352}{7}(24)

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{8448}{7}

\\ \sf\longmapsto TSA_{(New\:Cylinder)}=1204.7cm^2

So

\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{1204.7}{301.7}

\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{4}{1}

\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cylinder)}}:{TSA_{(Old\:Cylinder)}}=4:1}}}

Hence our correct option is Option C

6 0
2 years ago
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