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ivanzaharov [21]

3 years ago

6

400 times 45.2 Show all work

1 answer:

Bogdan [553]3 years ago

3 0

Answer: 18080!

Step-by-step explanation:400 x 450

1. Multiplying from the oneths place ( turning 45.2 into 450)

2x0=0 2x0=0 2x4=8 = So the first answer is 800

2. Multiply from the tenths place

5 x 0 = 0 5 x 0 = 0 5 x 4 = 20 = the second answer is 2000

Multiplying the hundredths place

4 x 0 = 0 4 x 0 = 0 4 x 4 = 16 = The third answer is 1600

800 + 2000 + 1600 = 18080 Hope this helped! :)

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The two functions below are used to represent the height of a rocket above the ground after it is launched. which rocket reaches

Bess [88]

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3 years ago

Manny is covering a book. The front of the book is 11.2 inches high and 7.3 inches wide. What is the area of the front of the bo

irina [24]

Answer:

81.76in^2

Step-by-step explanation:

Given data

Lenght of book= 11.2 inches

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3 years ago

Which symbol can be used to correctly compare the two fractions 2/5 and 35/100

Kaylis [27]

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Hope this helps :)

5 0

3 years ago

Read 2 more answers

use green's theorem to evaluate the line integral along the given positively oriented curve. c 9y3 dx − 9x3 dy, c is the circle

Rina8888 [55]

The line integral along the given positively oriented curve is -216π. Using green's theorem, the required value is calculated.

<h3>What is green's theorem?</h3>

The theorem states that,

$\int_CPdx+Qdy = \int\int_D(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy$

Where C is the curve.

<h3>Calculation:</h3>

The given line integral is

$\int_C9y^3dx-9x^3dy$

Where curve C is a circle x² + y² = 4;

Applying green's theorem,

P = 9y³; Q = -9x³

Then,

$\frac{\partial P}{\partial y} = \frac{\partial 9y^3}{\partial y} = 27y^2$

$\frac{\partial Q}{\partial x} = \frac{\partial -9x^3}{\partial x} = 27x^2$

$\int_C9y^3dx-9x^3dy = \int\int_D(-27x^2 - 27y^2)dx dy$

⇒ $-27\int\int_D(x^2 + y^2)dx dy$

Since it is given that the curve is a circle i.e., x² + y² = 2², then changing the limits as

0 ≤ r ≤ 2; and 0 ≤ θ ≤ 2π

Then the integral becomes

$-27\int\limits^{2\pi}_0\int\limits^2_0r^2. r dr d\theta$

⇒ $-27\int\limits^{2\pi}_0\int\limits^2_0 r^3dr d\theta$

⇒ $-27\int\limits^{2\pi}_0 (r^4/4)|_0^2 d\theta$

⇒ $-27\int\limits^{2\pi}_0 (16/4) d\theta$

⇒ $-108\int\limits^{2\pi}_0 d\theta$

⇒ $-108[2\pi - 0]$

⇒ -216π

Therefore, the required value is -216π.

Learn more about green's theorem here:

brainly.com/question/23265902

#SPJ4

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1 year ago

Backpack discounted 10% on sale for 37.80

Ksju [112]

I assume you want the original price?
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3 years ago