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4 years ago

A cereal box is 8 inches by 2 inches by 12 inches. What is its surface area?

Mathematics

2 answers:

dusya [7]4 years ago

4 0

Answer:

192 inches or 16 feet.

Step-by-step explanation:

Surface area is height x width x length, plug in your numbers are you get:

8 inches x 2 inches x 12 inches = 192 inches

kotegsom [21]4 years ago

3 0

Answer:

272 square inches

Step-by-step explanation:

Given: A cereal box is 8 inches by 2 inches by 12 inches.

To find: Surface area of the cereal box.

Solution:

The dimensions of the cereal box are 8 inches by 2 inches by 12 inches.

Let the length be 8 inches, width be 2 inches, and height be 12 inches.

Here, we need to find the total surface area of the cereal box.

The cereal box is in the shape of a rectangular prism.

The total surface area of the rectangular prism can be calculated as

$= 2(\text{length} \times \text{width} + \text{width} \times \text{height} + \text{length} \times \text{height})$

$= 2(\text{8} \times \text{2} + \text{2} \times \text{12} + \text{8} \times \text{12})$

$=2(16+24+96)$

$=2\times136$

$=272$

So, the surface area of the cereal box is 272 square inches.

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P(x) = x + 1x² – 34x + 343<br> d(x)= x + 9

Feliz [49]

Answer:

$x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343$

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > $\frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1$ . Go ahead and simplify.

$x=\frac{9}{d-1};\quad \:d\ne \:1$ .

Now, $\mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}$ .

$P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343$ .

Group the like terms... $1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343$ .

$\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1}$ > $1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343$ .

Now for $1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}$ .

$\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}$ .

Now for $33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}$ .

Thus we then get $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343$ .

Now we want to combine fractions. $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}$ .

$\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2$

$\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}$

$\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}$ .

Expand $81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac$ .

$-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1$ .

$\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}$

Therefore $P=\frac{-297d+378}{\left(d-1\right)^2}+343$ .

Hope this helps!

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Sever21 [200]

Answer:

Step-by-step explanation:

With those problems if you are not given a picture is good we draw one.

Because an angle bisector forms 2 congruent angles and because is given that < XVY ≅ < YVW then

m < XVY = m < YVW

2x+7 = x+15 , subtract x and 7 from both sides to isolate the like terms

2x-x = 15-7, combine like terms

x = 8

From the picture and the given we see that

m < XVW = m < XVY + m < YVW

m < XVW = 2x+7 + x+15 , combine like terms

m < XVW = 3x + 22, substitute x for 8

m < XVW = 3*8 + 22

m < XVW = 46

Check our work:

m < XVY = 2x+7 = 2*8 +7 = 16 +7 = 23

m < YVW = x+15 = 8 +15 = 23

m < XVW = m < XVY + m < YVW = 23+23 =46

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Answer:

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Step-by-step explanation:

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