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nika2105 [10]
3 years ago
11

A mini-cube measures 1.5 in. wide. A mega-cube is 5 times as wide. What is the volume of the mega-cube, rounded to the nearest t

enth?
Mathematics
2 answers:
Daniel [21]3 years ago
7 0

Answer:

the answer is  421.9 in³

Paul [167]3 years ago
3 0
My guess is it is round to  7.5 
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Pani-rosa [81]

Answer:

The answer is D

Step-by-step explanation: Trust me! hope this helps:)

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Un frasco de perfume tiene la capacidad de 1/20 de litro. ¿Cuántos frascos de perfume se pueden llenar con el contenido de una b
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15 frascos de perfume porque 3 / 4 = 15 / 20
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3 years ago
How do you determine if the line segments are congruent
Gemiola [76]
You must calculate the length of the segments.

Suppose segments AB and CD
Being A (m, n), B (p, q)  C(r,s) and D(t,u)
<span>
They are congruent if:

</span>\boxed{(p-m)^2+(q-n)^2=(t-r)^2+(u-s)^2}<span>

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3 years ago
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Follow the order of operations and evaluate these expressions. 1) 7+3(9-4)^2÷5, 12/3-4+7^2, (7-3)×3^3÷9, 5(7-3)^2÷(6-4)^3-9, 3×(
vovikov84 [41]

The solution to the first expression - 7+3(9-4)^2÷5 is given as 22.

To get the answer correctly, one must follow rudimentary rules of operations which are coined into the acronym BODMAS.

<h3>What is BODMAS?</h3>

This is the order in which mathematical operations must be executed.
B = Bracket

O = Orders (that is Powers, Indices or roots)

D= Division

M = Multiplication

A = Addition

S = Subtraction

Now lets see how we got 22 from the first set of operations:

<h3>Operation 1 (Example)</h3>

7+3(9-4)^2÷5 =

7+3 (5)^2÷5=

7+3 * 25÷5 =

7+3*5=

7+15=

22

Following the BODMAS rule and the example in Operation 1 above, we can state the remaining answers as follows:

<h3>

Operation 2</h3>

12/3-4+7^2 = 49

<h3 /><h3>Operation 3</h3>

(7-3)×3^3÷9 = 12

<h3>Operation 4</h3>

5(7-3)^2÷(6-4)^3-9 = 1

<h3>Operation 5</h3>

3×(7-5)^3÷(8÷4)^2-5 = 1

<h3>Operation 6</h3>

9+(3×10)/5×2-12​ = 9

See the link below for more about Mathematical Operations:

brainly.com/question/14133018

6 0
2 years ago
Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. Only 1 try
vagabundo [1.1K]

Using the shell method, the volume is

\displaystyle 2\pi \int_0^1 (2-x) \cdot 8x^3 \, dx = 16\pi \int_0^1 (2x^3 - x^4) \, dx

Each cylindrical shell has radius 2-x (the horizontal distance from the axis of revolution to the curve y=8x^3); has height 8x^3 (the vertical distance between a point on the x-axis in 0\le x\le1 and the curve y=8x^3).

Compute the integral.

\displaystyle 16 \pi \int_0^1 (2x^3 - x^4) \, dx = 16\pi \left(\frac{x^4}2 - \frac{x^5}5\right) \bigg|_{x=0}^{x=1} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 16\pi \left(\frac12 - \frac15\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{24}5\pi = \boxed{4.8\pi}

6 0
2 years ago
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