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Oksi-84 [34.3K]
2 years ago
5

The cereal box shown below is a rectangular prism.

Mathematics
1 answer:
joja [24]2 years ago
3 0

Answer:

Surface area is found:

Surface Area = 1700 cm²

Step-by-step explanation:

(The cereal box is shown in the ATTACHMENT)

The surface area of a rectangular prism can be found by added the areas of all 6 sides of the rectangular prism.

L = length = 20 cm

H = height = 30 cm

W = Width = 5 cm

Side 1:

A(1) = L×H

A(1) = 20×30

A(1) = 600 cm²

Side 2:

As the measurements of the side at the back of side 1 has the same measurement of side 1. then:

A(2) = 600 cm²

Side 3:

A(3) = L×W

A(3) = 20×5

A(3) = 100 cm²

Side 4:

As the measurements of the side at the back of side 4 has the same measurement of side 4. then:

A(4) = 100 cm²

Side 5:

A(5) = H×W

A(5) = 30×5

A(5) = 150 cm²

Side 6:

As the measurements of the side at the back of side 5 has the same measurement of side 5. then:

A(6) = 150 cm²

Surface Area:

Adding areas of all the sides

A(1) + A(2) + A(3) +A(4) + A(5) + A(6) = 600 + 600 + 100 +100 + 150 +150

Surface Area = 1700 cm²

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The volume of a cylinder is given by the formula , where r is the radius of the cylinder and h is the height. Which expression r
kotykmax [81]

The Volume of cylinder represented by the option B.

According to the statement

we have given that the volume formula of cylinder is V= (pi)r^2h, and we have to find that the which expression verify the volume formula for the cylinder given in diagram.

So,

We know that height of the cylinder is given by h = 2x + 7 and

radius r = x - 3.

We know that the formula of volume of cylinder is:

Volume of a cylinder = (pi)r^2h

and Substituting the given values in the above formula

And the volume becomes

Volume = (pi)r^2h

Volume = (pi)( x-3 )^2 (2x+7)

Volume = (pi) ( x^2 + 9 - 6x ) (2x+7)

Volume = (pi) ( 2x^3 + 7x^2 +18x +63 - 42x)

So, The Volume of cylinder represented by the option B.

Learn more about Volume here brainly.com/question/1972490

Disclaimer: This question is incomplete. Please find the full content below.

Question:

The volume of a cylinder is given by the formula V= pi^2h, where r is the radius of the cylinder and h is the height: which expression represents the volume of this cylinder?

a) Volume = (pi) ( 3x^3 + 7x^2 +14x +63 - 42x)

b) Volume = (pi) ( 2x^3 + 7x^2 +18x +63 - 42x)

c) Volume = (pi) ( 7x^3 + 7x^2 +11x +63 - 42x)

d) Volume = (pi) ( 11x^3 + 7x^2 +13x +63 - 42x)

#SPJ4

7 0
1 year ago
What is 20,164 ÷ 71<br><br><br> Answer Soon As Posible Please
Shkiper50 [21]
The answer of this question is 284
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3 years ago
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Ashley digs 7 holes.She puts 2 seeds in each hole.She has 3 seeds left over.How many seeds are there in all?
Semenov [28]
17 seeds are left over because in the 7 holes she puts 2 seeds in each which if you multiply it it will equal to 14 and plus the three that she has left over and if you add 14 and 3 you get 17 seeds. hope this helps!
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3 years ago
Read 2 more answers
Please help and give explanation
Firlakuza [10]

Answer:

100 mins

Step-by-step explanation:

total no of gallons pool can hold=4500

gallons pool already holds=1500

rate of adding water= 30 gallons per min

time he can continue filing= 4500-1500/30

=3000/30

=100 minutes (m value)

correct me if I'm wrong appreciate me if I'm good

8 0
3 years ago
Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d
aliya0001 [1]

The Lagrangian

L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)

has critical points where the first derivatives vanish:

L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}

L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}

L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}

L_\lambda=x^4+y^4+z^4-13=0

We can't have x=y=z=0, since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of \pm\sqrt[4]{13}. For example, if y=z=0, then x^4=13\implies x=\pm\sqrt[4]{13}.

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find \lambda=-\frac1{\sqrt{26}} (taking the negative root because x^2,y^2,z^2 must be non-negative), and we can immediately find the critical points from there. For example, if z=0, then x^4+y^4=13. If both x,y are non-zero, then x^2=y^2=-\frac1{2\lambda}, and

xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}

\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}

and for either choice of x, we can independently choose from y=\pm\sqrt[4]{\frac{13}2}.

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then x^2=y^2=z^2=-\frac1{2\lambda}. We have

xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}

\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}

and similary y,z have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate f at each critical point; you should end up with a maximum value of \sqrt{39} and a minimum value of \sqrt{13} (both occurring at various critical points).

Here's a comprehensive list of all the critical points we found:

(\sqrt[4]{13},0,0)

(-\sqrt[4]{13},0,0)

(0,\sqrt[4]{13},0)

(0,-\sqrt[4]{13},0)

(0,0,\sqrt[4]{13})

(0,0,-\sqrt[4]{13})

\left(\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)

\left(\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)

\left(-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)

\left(-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)

\left(\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)

\left(\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)

\left(-\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)

\left(-\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)

\left(0,\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)

\left(0,\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)

\left(0,-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)

\left(0,-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)

\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)

\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)

5 0
3 years ago
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