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jolli1 [7]
3 years ago
15

What is the perimeter??

Mathematics
1 answer:
nalin [4]3 years ago
6 0

Answer:

soma tudo se n for multiplica

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If 2x = 5, 3y = 4, and 4z = 3, what is the value of 24xyz ?
NISA [10]
2x=5 so 5÷2 = x (2.5)
3y = 4 so 4÷3 = y (1.3 recurring)
4z = 3 so 3÷4 = z (0.75)

implement: 24x2.5x1.3•x0.75

hope i answered right
3 0
4 years ago
What would the measure of DEC be
IceJOKER [234]
So if <AED =100 degrees result <BEC = 100 degrees too and 360-200=160
so 160/2 = 80 what mean that angle DEC=80 degrees 

hope helped 
3 0
3 years ago
Explain in words how to solve the following problem.Maske sure to include all steps and the answer. 3/7 divided by 1/2
Maksim231197 [3]
To divide by a fraction is to multiply by its recipricle.
3/7 × 2/1
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6/7
3 0
3 years ago
Read 2 more answers
Compute 1 + 2 + 3 +....+ 1,997 + 1,998 + 1,999
sp2606 [1]

Answer:

1 999 000

Step-by-step explanation:

Formula:

1+2+3+.\ .\ .+n=\frac{n\times \left( n+1\right)  }{2}

………………………………………

Then

1+2+3+....+1997+1998+1999=\frac{1999\times \left( 1999+1\right)  }{2}

1+2+3+....+1997+1998+1999=\frac{1999\times \left( 2000\right)  }{2}

1+2+3+....+1997+1998+1999=\frac{3998000  }{2}

1+2+3+....+1997+1998+1999=1999000

7 0
2 years ago
A piece of cardboard is 13 inches by 26 inches. A square is to be cut from each corner and the sides folded up to make an open-t
Vanyuwa [196]

Answer:

Hence the maximum possible volume will be the 778.53 c.c

Step-by-step explanation:

Given:

A rectangle with 13 x 26 dimensions

And corners are cut to form side squares.

To Find:

Maximum possible volume for box

Solution :

Consider a rectangle of 13 x 26 dimension with and side of square  at corner be x.

(Refer the attachment)

Now,

Formulating the volume equation for the box

So corner square sides we are going to fold up which makes height of the box

and remaining part will be length and breadth

As shown in fig,

Length=26-x

breadth=13-x

And height will be x

V(x)=x*(26-x)*(13-x)

To get maximum volume differentiate the above equation,

V(x)=x*(26*13-26*x-13*x+x^2)

V(x)=x^3-39x^2+338x\\

V'(x)=3x^2-78x+338

V''(x)=6x-78

Now ,Solve the Quadratic Equation to get x values,

3x^2-78x+338=0

x=[-b±(b^2-4ac)^1/2]/2a

x=[78±Sqrt[(78)^2-4*338*3)]/2*3

x=[78±Sqrt(3028)]/6

x=[78±55.027]/6

x=78+55.027/6 or x=78-55.027/6

x=22.17  or x=3.8288

Use these values in 6x-78 to know which value posses the max and min value for the function.

So when x=22.17

6x-78=6*22.17-78

=55.02>0  i.e function will have minimum value .

When x=3.8288

6*3.8288-78

=-55.0272<0 i.e. Function will have maximum value

Now, the function will defines the maximum volume

V(x)=x^3-39x^2+338x

V(x)=3.8288^3-39*(3.82883)^2+338*3.8288

V(x)=56.13-571.73+1294.13

V(x)=778.53 C.C

6 0
3 years ago
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