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3 years ago

Calculate the area of the irregular polygon shown below:

Mathematics

1 answer:

nlexa [21]3 years ago

3 0

The answer is 44 in.²

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What is (-6-8i)-(-9+4i) in standard form?

Ray Of Light [21]

Answer:

3 -12i

Step-by-step explanation:

(-6-8i)-(-9+4i)

distribute the minus sign

-6 -8i +9 -4i

combine like terms

-6+9 -8i -4i

3 -12i

8 0

3 years ago

Help me ;-;<br><br> Find (2)(‐3)

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2 years ago

Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.

rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed as:

$y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}$

Thus, the area of the surface is as follows:

$\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA$

$= \iint_D \sqrt{x^2 +y^2 +1 } \ dA$

$= \int^{2 \pi}_{0} \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr \ d \theta$

$=2 \pi \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr$

$= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0$

$= 2 \pi \times \dfrac{1}{3} \Bigg [ (37)^{3/2} - 1 \Bigg]$

$= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]$

3 0

2 years ago

The product of x and y is less than or equal to 4.

timama [110]

Answer:

xy (less than or equal to sign) 4

5 0

3 years ago

PLZ HELP I WILL GIVE BRAINLIEST...

Delicious77 [7]

Answer:

a. 6

b. 7

c. (b) Kira's

why? b. it has the large sample size

Step-by-step explanation:

7 0

3 years ago