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3 years ago

13. Fill in this diagram so that each of the rows, columns and

Mathematics

1 answer:

Natasha2012 [34]3 years ago

6 0

Answer:

D. 24

Step-by-step explanation:

Let's name the diagram

Row 1: a1, a2, a3

Row 2: b1, b2, b3

Row 3: c1, c2, c3

From the diagram,

b2 and c2 have been given

All rows, columns and diagonal must sum up to 18

If b2=6 and c2=4

a2=18-(6+4)

=18-10

a2=8

Assume a1=4

Diagonals a1+b3+c3=18

4+6+c3=18

10+c3=18

c3=18-10

Assume a3=6

Diagonals a3+b2+c1=18

6+6+c1=18

12+c1=18

c1=18-12

So b1=18-(6+4)

=18-10

b1=8

b3=18-(6+8)

=18-14

b3=4

Input all the numbers into the boxes

We have,

4 8 6

8 6 4

6 4 8

Corner numbers are a1,a3,c1,c3

=4,6,6,8

Sum of all the corner numbers=4+6+6+8

=24

D. 24

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the area of a window is 40 square feet. the length is the square root of 20 feet. what is the width of the window

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----------------------------------------------------
Given information
----------------------------------------------------
Area = 40ft²
Length = √20 ft

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Formula
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Area = Length x Width

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Find Width
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40 = √20 x width
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3 years ago

cylinder shaped can needs to be constructed to hold 200 cubic centimeters of soup. The material for the sides of the can costs 0

Dafna11 [192]

Answer:

Radius=2.09 cm

Height,h=14.57 cm

Step-by-step explanation:

We are given that

Volume of cylinderical shaped can=200 cubic cm.

Cost of sides of can=0.02 cents per square cm

Cost of top and bottom of the can =0.07 cents per square cm

Curved surface area of cylinder= $2\pi rh$

Area of circular base=Area of circular top= $\pi r^2$

Total cost,C(r)= $0.02\times 2\pi rh+2\pi r^2\times 0.07$

Volume of cylinder, $V=\pi r^2 h$

$200=\pi r^2 h$

$h=\frac{200}{\pi r^2}$

Substitute the value of h

$C(r)=0.02\times 2\pi r\times \frac{200}{\pi r^2}+2\pi r^2\times 0.07$

$C(r)=\frac{8}{r}+0.14\pi r^2$

Differentiate w.r.t r

$C'(r)=-\frac{8}{r^2}+0.28\pi r$

$C'(r)=0$

$-\frac{8}{r^2}+0.28\pi r=0$

$0.28\pi r=\frac{8}{r^2}$

$r^3=\frac{8}{0.28\pi}=9.095$

$r=(9.095)^{\frac{1}{3}}=2.09$

Again, differentiate w.r.t r

$C''(r)=\frac{16}{r^3}+0.28\pi$

Substitute the value of r

$C''(2.09)=\frac{16}{(2.09)^3}+0.28\pi=2.63>0$

Therefore,the product cost is minimum at r=2.09

h= $\frac{200}{\pi (2.09)^2}=14.57$

Radius of can,r=2.09 cm

Height of cone,h=14.57 cm

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3 years ago

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