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3 years ago

Help me understand please

Mathematics

1 answer:

erastova [34]3 years ago

6 0

Answer:

V=pi x (r)^2 x h

V-volume r-radius h-height

V= pi x (3)^2 x 9

V= pi x 9 x 9

V= pi x 81

V= 254.47 (rounded)

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Use the inequality 24 ≥ 58 + 5(x - 3.8)

spayn [35]

Answer:

Work shown below!

Step-by-step explanation:

24 ≥ 58 + 5(x - 3.8)

Distribute;

24 ≥ 58 + 5x - 19

Collect like terms;

24 ≥ 39 + 5x

Subtract 39 from both sides;

-15 ≥ 5x

Divide both sides by 5;

x ≥ -3

4 0

3 years ago

Please help me in Geometry!

Masja [62]

Answer:

You get an perpendicular line through given point.

Step-by-step explanation:

In this method of drawing perpendicular line through given point,

you will draw an arc from given point R, so as to cut given line at A and B.

now ,length RA will be equal to length RB (radius of circle)

now, with length RA you will place compass at A and B to cut the arcs at point P (let) on other side of given line.Now, join PA and PB.

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8 0

3 years ago

Read 2 more answers

HELP ME PLEASE I BEG YOU!!!

antiseptic1488 [7]

Slope is negative for each...Remember Y-int and rise/run for slope.
Y=mx+b

Y= -2/3X - 2

Y= -5/2x + 2

7 0

3 years ago

The Candle Factory is producing a new candle. It has a radius of 3 inches and a height of 5 inches. What is the surface area of

garik1379 [7]

Answer:

87.92 square inches

Step-by-step explanation:

Surface area for cylinder is given by $2\pi r^2 + 2\pi rh$

where r is the radius and h is the height of cylinder.

Given that candle is cylindrical in shape

Thus, Surface area for candle is calculated as given below\\

$Surface / area \ for \ candle = 2\pi r^2 + 2\pi rh\\=>Surface / area \ for \ candle = 2*3.14*3^2 + 2*3.14*5\\=>Surface / area \ for \ candle = 2*3.14*3^2 + 2*3.14*5 = 56.52 + 31.4\\=>Surface / area \ for \ candle = 56.52 + 31.4 = 87.92$

Thus, total surface area for candle is 87.92 square inches.

8 0

3 years ago

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

4 0

3 years ago

Help me understand please​

Help me understand please