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3 years ago

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Mathematics

2 answers:

Wewaii [24]3 years ago

6 0

Answer:

10 × 5 - (- 6)² + (- 8) = 6

(- 5)² × 3 ÷ 5 + 9 = 24

(10 ÷ (- 5) - (- 2)) × (- 3)³ = 0

4 × (- 6) ÷ 8 + 3³ = 24

(- 5)² - 2 × (- 9) + 6 = 49

3 × 10 + 8 - 4² = 22

(- 9) - (- 8) + 2 × 4² = 31

(- 3)³ - 2 + 8 ÷ (- 8) = - 30

8 ÷ (- 4) × (- 6)² + 7 = - 65

4 × (- 8) + 6 - (- 2)³ = - 18

Step-by-step explanation:

10 × 5 - (- 6)² + (- 8)

10 × 5 - 36 - 8

50 - 36 - 8

50 - 44

(- 5)² × 3 ÷ 5 + 9

25 × 3 ÷ 5 + 9

75 ÷ 5 + 9

15 + 9

(10 ÷ (- 5) - (- 2)) × (- 3)³

(10 ÷ (- 5) - (- 2)) × - 27

(10 ÷ - 5 + 2) × - 27

(- 2 + 2) × - 27

0 × - 27

4 × (- 6) ÷ 8 + 3³

4 × (- 6) ÷ 8 + 27

- 24 ÷ 8 + 27

- 3 + 27

(- 5)² - 2 × (- 9) + 6

25 - 2 × - 9 + 6

25 + 18 + 6

3 × 10 + 8 - 4²

3 × 10 + 8 - 16

30 + 8 - 16

30 - 8

(- 9) - (- 8) + 2 × 4²

(- 9) - (- 8) + 2 × 16

- 9 + 8 + 2 × 16

- 9 + 8 + 32

32 - 1

(- 3)³ - 2 + 8 ÷ (- 8)

- 27 - 2 + 8 ÷ - 8

- 27 - 2 - 1

- 27 - 3

- 30

8 ÷ (- 4) × (- 6)² + 7

8 ÷ (- 4) × 36 + 7

- 2 × 36 + 7

- 72 + 7

- 65

4 × (- 8) + 6 - (- 2)³

4 × (- 8) + 6 - (- 8)

- 32 + 6 + 8

- 32 + 14

- 18

Vilka [71]3 years ago

6 0

$10 x 5-(-6)^2 +(-8)\\10x5-36+(-8)\\50-36+(-8)\\50-28\\22$

$(-5)^2$ x3÷5+9

25x3÷5+9

75÷5+9

15+9

(10÷(-5)-(-2))x(-3)³

((-2)-(-2))x(-3)³

0x(-3)³

0x(-27)

4x(-6)÷8+3³

4x(-6)÷8+27

(-24)÷8+27

(-3)+27

(-5)²-2x(-9)+6

25-2x(-9)+6

25+18+6

43+6

3x10+8-4²

3x10+8-16

30+8-16

38-16

(-9)-(-8)+2x4²

(-9)-(-8)+2x16

(-9)-(-8)+32

(-9)-24

(-33)

(-3)³-2+8÷(-8)

(-27)-2+8÷(-8)

(-27-2+(-1)

(-27)-1

(-28)

8÷(-4)x(-6)²+7

8÷(-4)x(-36)+7

8÷144+7

0.055+7

7.055

4x(-8)+6-(-2)³

4x(-8)+6-8

(-32)+6-8

(-26)-8

(-34)

I hope this will help you. ^^

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son4ous [18]

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Step-by-step explanation:

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3 years ago

Electric charge is distributed over the disk x2 + y2 ≤ 16 so that the charge density at (x, y) is rho(x, y) = 2x + 2y + 2x2 + 2y

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Answer:

Required total charge is $256\pi$ coulombs per square meter.

Step-by-step explanation:

Given electric charge is dristributed over the disk,

$x^2=y^2\leq 16$ so that the charge density at (x,y) is,

$\rho (x,y)=2x+2y+2x^2+2y^2$

To find total charge on the disk let Q be the total charge and $x=r\cos\theta,y=r\sin\theta$ so that,

$Q={\int\int}_Q\rho(x,y) dA$ where A is the surface of disk.

$=\int_{0}^{2\pi}\int_{0}^{4}(2x+2y+2x^2+2y^2)dA$

$=\int_{0}^{2\pi}\int_{0}^{4}(2r\cos\theta+2r\sin\theta+2r^2 \cos^{2}\theta+2r^2\sin^2\theta)rdrd\theta$

$=2\int_{0}^{2\pi}\int_{0}^{4}r^2(\cos\theta+\sin\theta)drd\theta+2\int_{0}^{2\pi}\int_{0}^{4}r^3drd\theta$

$=\frac{2}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)\Big[r^3\Big]_{0}^{4}d\theta+2\int_{0}^{2\pi}\Big[\frac{r^4}{4}\Big]d\theta$

$=\frac{128}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)d\theta+128\int_{0}^{2\pi}d\theta$

$=\frac{128}{3}\Big[\sin\theta-\cos\theta\Big]_{0}^{2\pi}+128\times 2\pi$

$=\frac{128}{3}\Big[\sin 2\pi-\cos 2\pi-\sin 0+\cos 0\Big]+256\pi$

$=256\pi$

Hence total charge is $256\pi$ coulombs per square meter.

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3 years ago