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3 years ago

How much cardboard is needed to make the tissue box above? Square inches 8 3 3

Mathematics

1 answer:

Katen [24]3 years ago

5 0

The amount of cardboard needed to make a cuboid box of dimensions 8 inch, 3 inches and 3 inches is 114 sq. inches.

<h3>What is the surface area of cuboid?</h3>

Let the three dimensions(height, length, width) be x, y,z units respectively.

The surface area of the cuboid is given by

$S = 2(a\times b + b\times c + c\times a)$

For this case, tissue box is almost cuboid shaped.

Also, its dimensions are given being 8 inches, 3 inches and 3 inches.

Suppose we measure the amount of cardboard needed in terms of area, then, the amount of cardboard needed to make that box(without any whole, full cuboid) is equal to the area of its surface(either outer or inner if we assume 0 inches thickness of cardboard),

Thus, we get:

Amount of cardboard needed = surface area of cuboid box with dimensions 8 by 3 by 3 (in inches)

= $2(8 \times 3 + 3 \times 3 + 3 \times 8) = 114 \: \rm in^2$

Thus, the amount of cardboard needed to make a cuboid box of dimensions 8 inch, 3 inches and 3 inches is 114 sq. inches.

Learn more about surface area of cuboid here:

brainly.com/question/13522634

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Task 1:

laiz [17]

Answer:

The toy's maximum height is $10.12\ \text{feet}$ .

Step-by-step explanation:

The equation that models the toy's height in feet from the ground at t seconds after he threw it is given by :

$h = -2t^2 + 7t + 4$ .....(1)

It is required to find the toy's maximum height. For maxima or minima, put

$\dfrac{dh}{dt}=0$

Plugging the value of t in above equation,

$\dfrac{d(-2t^2 + 7t + 4)}{dt}=0\\\\-4t+7=0\\\\t=\dfrac{7}{4}\\\\t=1.75\ s$

Now put t = 1.75 s in equation (1) such that,

$h = -2(1.75)^2 + 7(1.75) + 4\\\\h=10.12\ \text{feet}$

So, the toy's maximum height is $10.12\ \text{feet}$ .

4 0

4 years ago

h(x)=x 2 −1h, left parenthesis, x, right parenthesis, equals, x, squared, minus, 1 Over which interval does h hh have a negative

Semenov [28]

Answer:

(C)−3 ≤ x ≤ 1

Step-by-step explanation:

The average rate of change of function h over the interval $a \leq x\leq b$ , is given by this expression:

$\dfrac{h(b)-h(a)}{b-a}$

Given the function $h(x)=x^2-1$ on the interval:− 3 ≤ x ≤ 1

$h(1)=1^2-1=0\\h(-3)=(-3)^2-1=9-1=8$

The average rate of change:

$\dfrac{h(b)-h(a)}{b-a}=\dfrac{0-8}{1-(-3)}=\dfrac{-8}{4}=-2$

Therefore, the function has a negative average rate of change over the interval − 3 ≤ x ≤ 1.

CHECK:

(A)Average rate of change of h(x) over the interval − 3 ≤ x ≤ 5=2

(B)Average rate of change of h(x) over the interval 1 ≤ x ≤ 4=5

(D)Average rate of change of h(x) over the interval − 1 ≤ x ≤ 5=4

5 0

3 years ago

2.3+0.02(x+20)-4.8=-9<br><br> Solve for x

Naya [18.7K]

Answer:

$x=-345$

Step-by-step explanation:

$2.3+0.02\left(x+20\right)-4.8=-9$

$\mathrm{Multiply\:both\:sides\:by\:}100$

$2.3\cdot \:100+0.02\left(x+20\right)\cdot \:100-4.8\cdot \:100=-9\cdot \:100$

$230+2\left(x+20\right)-480=-900$

$\mathrm{Subtract\:}230-480\mathrm{\:from\:both\:sides}$

$230+2\left(x+20\right)-480-\left(230-480\right)=-900-\left(230-480\right)$

$2\left(x+20\right)=-650$

$\mathrm{Divide\:both\:sides\:by\:}2$

$\frac{2\left(x+20\right)}{2}=\frac{-650}{2}$

$x+20=-325$

$\mathrm{Subtract\:}20\mathrm{\:from\:both\:sides}$

$x+20-20=-325-20$

$\bold{x=-345}$

5 0

3 years ago

How many solutions does this equation have?

maxonik [38]

Answer:

No solutions

Step-by-step explanation:

The lines are parallel.

5 0

3 years ago

Read 2 more answers

Please help me with this!!

Dennis_Churaev [7]

angle QRP=180-(64+47)

=69°

Using sin rule,

p/sinP=q/sinQ.

7.6/sin64=q/sin47

q=6.2cm

Using the formula;

area=1/2p*qsinR

we have;

1/2×7.6×6.2×sin69°=21.995

answer= 22

7 0

3 years ago

How much cardboard is needed to make the tissue box above? Square inches 8 3 3￼

How much cardboard is needed to make the tissue box above? Square inches 8 3 3