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bonufazy [111]
2 years ago
9

Luis has a pyramid shaped plant pot. It has a square base with a side length of

Mathematics
1 answer:
pychu [463]2 years ago
6 0

Answer:

The soil will occupy in 11664cm³ of the pot which is 3/4 or 0.75 of the squared based pyramid pot.

Step-by-step explanation:

The volume of a square based pyramid = V = a²h/3

a = base edge or side length = 36cm

h = height = 36cm

V = 36² × 36/3

V = 15552cm³

Luis wants to fill the pot with soil so that the soil takes up 75 % the pot's volume.

75% of the pot's volume =

75/100 × 15552cm³

= 11664cm³

The soil will occupy 11664cm³ of the pot which is 3/4 or 0.75 of the squared based pyramid pot.

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Solve y=4x+rx+6 for x.
MariettaO [177]

Answer:

x = y−6/r+4

Step-by-step explanation:

Let's solve for x.

y=4x+rx+6

Step 1: Flip the equation.

rx+4x+6=y

Step 2: Add -6 to both sides.

rx+4x+6+−6=y+−6

rx+4x=y−6

Step 3: Factor out variable x.

x(r+4)=y−6

Step 4: Divide both sides by r+4.

x(r+4)/r+4=y−6/r+4

5 0
2 years ago
Decide, without calculation, if each of the integrals below are positive, negative, or zero. Let D be the region inside the unit
Schach [20]

The integrals over B and T will be positive. Keeping y fixed, xe^x is strictly increasing over D as x increases, so the integrals over x (i.e. the bottom/top left quadrants of D) is negative but the integrals over x>0 are *more* positive.

The integrals over R and L are zero. If we take f(x,y)=xe^x, then f(x,-y)=f(x,y), which is to say f is symmetric across the x-axis. For the same reason, the integral over all of D is also zero.

4 0
2 years ago
The sum of three consecutive integers is 192. what are the integers
iVinArrow [24]

Answer:

63, 64, 65

Step-by-step explanation:

Consecutive integers have a difference of 1 between them

let the 3 integers be n, n + 1 and n + 2 , then their sum equals 192

n + n + 1 + n + 2 = 192

3n + 3 = 192 ( subtract 3 from both sides )

3n = 189 ( divide both sides by 3 )

n = 63 ⇒ n + 1 = 64 and n + 2 = 65

the 3 integers are 63, 64 and 65


4 0
2 years ago
Read 2 more answers
What is true about the completely simplified sum of the polynomials 3x2y2 − 2xy5 and −3x2y2 + 3x4y?
Assoli18 [71]

Answer:

The sum is a binomial with a degree of 6

Step-by-step explanation:

we have

(3x^{2}y^{2}-2xy^{5})+(-3x^{2}y^{2}+3x^{4}y)

Group terms that contain the same variable

(3x^{2}y^{2}-3x^{2}y^{2})-2xy^{5}+3x^{4}y

0-2xy^{5}+3x^{4}y

-2xy^{5}+3x^{4}y

The sum is a binomial ( two terms) with a degree of 6  

-2xy^{5} has a degree of 6 (x has an exponent of 1, y has 5, and 1+5=6)

6 0
2 years ago
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Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 1 + sec
masha68 [24]

Using the washer method, the volume is given by the integral

\displaystyle\pi\int_{-\pi/3}^{\pi/3}\bigg((3-1)^2-((1+\sec x)-1)^2\bigg)\,\mathrm dx=2\pi\int_0^{\pi/3}(4-\sec^2x)\,\mathrm dx

where 3 - 1 = 2 is the distance from <em>y</em> = 3 to the axis of revolution, and similarly (1 + sec(<em>x</em>)) - 1 = sec(<em>x</em>) is the distance from <em>y</em> = 1 + sec(<em>x</em>) to the axis. The integrand is symmetric about <em>x</em> = 0, so the integral "folds" in on itself, and the integral from -π/3 to π/3 is twice the integral from 0 to π/3.

So the volume is

\displaystyle2\pi\int_0^{\pi/3}(4-\sec^2x)\,\mathrm dx=2\pi(4x-\tan x)\bigg|_0^{\pi/3}=\boxed{\dfrac{8\pi^2}3-2\pi\sqrt3}

4 0
2 years ago
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