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Alex_Xolod [135]
2 years ago
10

Ricky is filling an 8 inch by 4 1/2 inch rectangular box with packing peanuts. The peanuts are cubes that come in two different

sizes ,1 cubic inch and 1/8 inch .How many peanuts of each size are needed to fill the box?
Mathematics
1 answer:
Marina86 [1]2 years ago
6 0
Ok, so we can fill the part of the box that corresponds to the 8x4 inches with the boxes that are 1 cubic inch - that wound be 8x4, that is 32 such boxes.

then the rest would be 8x \frac{1}{2} inches, that is 4 cubic inches.

In order to fill in 4 cubic inches with  \frac{1}{8} inches boxes we calculate:

4/ \frac{1}{8}=4*8=32  So it's also 32 such small boxes!

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What the answer please help marking brainliest if it's correct and explain​
kolezko [41]

Answer: The answer is C or Median

Step-by-step explanation:

8 0
2 years ago
How does the volume of the cylinder compare to the volume of the cone
aniked [119]
A cylinder with the same radius and height as a cone will have 3 times the volume of the cone.

Vcone = (1/3)*π*r^2*h
Vcylinder = π*r^2*h
5 0
2 years ago
Suppose Upper F Superscript prime Baseline left-parenthesis x right-parenthesis equals 3 x Superscript 2 Baseline plus 7 and Upp
Sedaia [141]

It looks like you're given

<em>F'(x)</em> = 3<em>x</em>² + 7

and

<em>F</em> (0) = 5

and you're asked to find <em>F(b)</em> for the values of <em>b</em> in the list {0, 0.1, 0.2, 0.5, 2.0}.

The first is done for you, <em>F</em> (0) = 5.

For the remaining <em>b</em>, you can solve for <em>F(x)</em> exactly by using the fundamental theorem of calculus:

F(x)=F(0)+\displaystyle\int_0^x F'(t)\,\mathrm dt

F(x)=5+\displaystyle\int_0^x(3t^2+7)\,\mathrm dt

F(x)=5+(t^3+7t)\bigg|_0^x

F(x)=5+x^3+7x

Then <em>F</em> (0.1) = 5.701, <em>F</em> (0.2) = 6.408, <em>F</em> (0.5) = 8.625, and <em>F</em> (2.0) = 27.

On the other hand, if you're expected to <em>approximate</em> <em>F</em> at the given <em>b</em>, you can use the linear approximation to <em>F(x)</em> around <em>x</em> = 0, which is

<em>F(x)</em> ≈ <em>L(x)</em> = <em>F</em> (0) + <em>F'</em> (0) (<em>x</em> - 0) = 5 + 7<em>x</em>

Then <em>F</em> (0) = 5, <em>F</em> (0.1) ≈ 5.7, <em>F</em> (0.2) ≈ 6.4, <em>F</em> (0.5) ≈ 8.5, and <em>F</em> (2.0) ≈ 19. Notice how the error gets larger the further away <em>b </em>gets from 0.

A <em>better</em> numerical method would be Euler's method. Given <em>F'(x)</em>, we iteratively use the linear approximation at successive points to get closer approximations to the actual values of <em>F(x)</em>.

Let <em>y(x)</em> = <em>F(x)</em>. Starting with <em>x</em>₀ = 0 and <em>y</em>₀ = <em>F(x</em>₀<em>)</em> = 5, we have

<em>x</em>₁ = <em>x</em>₀ + 0.1 = 0.1

<em>y</em>₁ = <em>y</em>₀ + <em>F'(x</em>₀<em>)</em> (<em>x</em>₁ - <em>x</em>₀) = 5 + 7 (0.1 - 0)   →   <em>F</em> (0.1) ≈ 5.7

<em>x</em>₂ = <em>x</em>₁ + 0.1 = 0.2

<em>y</em>₂ = <em>y</em>₁ + <em>F'(x</em>₁<em>)</em> (<em>x</em>₂ - <em>x</em>₁) = 5.7 + 7.03 (0.2 - 0.1)   →   <em>F</em> (0.2) ≈ 6.403

<em>x</em>₃ = <em>x</em>₂ + 0.3 = 0.5

<em>y</em>₃ = <em>y</em>₂ + <em>F'(x</em>₂<em>)</em> (<em>x</em>₃ - <em>x</em>₂) = 6.403 + 7.12 (0.5 - 0.2)   →   <em>F</em> (0.5) ≈ 8.539

<em>x</em>₄ = <em>x</em>₃ + 1.5 = 2.0

<em>y</em>₄ = <em>y</em>₃ + <em>F'(x</em>₃<em>)</em> (<em>x</em>₄ - <em>x</em>₃) = 8.539 + 7.75 (2.0 - 0.5)   →   <em>F</em> (2.0) ≈ 20.164

4 0
2 years ago
the average age of four men is 40. If the sum of the ages of three of them is 115, what's the age of the fourth man?​
pashok25 [27]

Answer:

35

Step-by-step explanation:

if 1 man is 37

and 1 man is 43

the 3rd man is 45

they add to 115

to average 40, we gotta add 35

this old mans giving u that!

3 0
2 years ago
Which sign makes the statement true? 93/100 ? 2/5
polet [3.4K]

Answer:

>.

Step-by-step explanation:

93/100 = 0.93

2/5 = 0.4,  so:

93/100 > 2/5.

7 0
3 years ago
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