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3 years ago

Diameter of a circular garden pool is six feet. what is the area of the pool in algebra

Mathematics

1 answer:

Art [367]3 years ago

4 0

For this case we have that by definition, the area of a circle is given by:

$A = \pi * r ^ 2$

Where:

r: It is the radius of the circle

As data we have that $r = 6 \ ft$

Then, replacing we have:

$A = \pi * 6 ^ 2\\A = 36 \pi$

Taking $\pi = 3.14$

A = 113.04

So, the pool area is $113.04 \ ft ^ 2$

Answer:

$113.04 \ ft ^ 2$

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AURORKA [14]

Answer:

2x+50 and 5x-55 both are congruent or have same measure.

Step-by-step explanation:

Since we want to prove that both lines are parallel, this means no theorems that involve with parallel lines apply here.

First of, we know that AC is a straight line and has a measure as 180° via straight angle.

x+25 and 2x+50 are supplementary which means they both add up to 180°.

Sum of two measures form a straight line which has 180°.

Therefore:-

x+25+2x+50=180

Combine like terms:-

3x+75=180

Subtract 75 both sides:-

3x+75-75=180-75

3x=105

Divide both sides by 3.

x=35°

Thus, x = 35°

Then we substitute x = 35 in every angles/measures.

x+25 = 35°+25° = 60°

2x+50 = 2(35°)+50° = 70°+50° = 120°

5x-55 = 5(35°)-55 = 175°-55° = 120°

Since 2x+50 and 5x-55 have same measure or are congruent, this proves that both lines are parallel.

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2 years ago

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14, 10, 2, 5, 8, 12, 12, 3, 2, 2, 13, 15, 6 Find the median

hoa [83]

Answer:

Step-by-step explanation:

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3 years ago

The line integral of (2x+9z) ds where the curve is given by the parametric equations x=t, y=t^2, z=t^3 for t between 0 and 1. Pl

Naya [18.7K]

Let r = (t,t^2,t^3)

Then r' = (1, 2t, 3t^2)

General Line integral is:
$\int_a^b f(r) |r'| dt$

The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector $\sqrt{(x')^2 + (y')^2 + (z')^2}$

$\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt$

Fortunately, this simplifies nicely with a 'u' substitution.

Let u = 1+4t^2 +9t^4

du = 8t + 36t^3 dt

$\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u} du \\ \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u} du \\ \\ \frac{1}{4} \int_0^1 \sqrt{u} du$

After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
$=\frac{1}{4} |_0^1 (\frac{2}{3}) (1+4t^2 +9t^4)^{3/2} \\ \\ =\frac{1}{6} (14^{3/2} - 1)$

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3 years ago

sebuah topi berbentuk kerucut dengan panjang diameter 18 cm dan tinggi topi tersebut 20cm berapakah volume nya

madreJ [45]

Answer:

The area of the cone is "1,695.6 cm³".

Step-by-step explanation:

The given values are:

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d = 18 cm

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