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3 years ago

Please help! multiplying and dividing integers no link pls

Mathematics

2 answers:

frosja888 [35]3 years ago

7 0

Answer:

Below,... ;->

Step-by-step explanation:

With Multiplication and Division you just carry out normal Multiplication and Division. If the equation has an EVEN number of negative signs the answer will always be positive (the sign cancel each other out). In turn if there is an ODD number of negative signs the answer will always be negative,... (with Addition and Subtraction it is different but with Multiplication and Division working out problems works in this way,...)

1 = 96 (even number of negative signs,... (in the whole equation))

2 = -4 (odd number of negative signs)

3 = -60

4 = -176

5 = 44

6 = -76

7 = -64

8 = 306

9 = 64

10 = 48

11 = 24

12 = 48

13 = -7 (do the division like you normally would except now just add the neg sign when appropriate (in this case there is 1 neg sign so answer is neg))

14 = 16

15 = -7

16 = 2

17 = -4

18 = 13

19 = 5 (do the parenthesis first (order of operations))

20 = 5 (do the parenthesis first (order of operations))

OKAY,... I think that does it,... Best of luck! Hope that helps,... Chow,...!

posledela3 years ago

4 0

Answer:

1(96) 2(-4) 3(-60) 4(-176) 5(44) 6(-76) 7(-64) 8(306) 9(64) 10(48) 11(24) 12(48) 13(-7) 14(16) 15(-7) 16(2) 17(-4) 18(38) 19(5) 20(5)

Step-by-step explanation:

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fgiga [73]

Answer:

$|z| = 27$ -- Modulus

$\theta = \frac{\pi}{3}$ --- Argument

Step-by-step explanation:

Given

$((\sqrt 3)(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18}))^6$

Required

Determine the modulus and the argument

We have that:

$z = ((\sqrt 3)(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18}))^6$

Expand:

$z = (\sqrt 3)^6(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})^6$

$z = 27(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})^6$

A complex equation can be expressed as:

$z = |z| e^{i\theta}$

Where

$|z| = modulus$

$\theta = argument$

Where

$e^{i\theta} = (cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})$

So:

$z = 27(cos\frac{\pi}{18} + i\ sin\frac{\pi}{18})^6$ becomes

$z = 27(e^{i\frac{\pi}{18}})^6$

By comparison:

$e^{i\theta} = (e^{i\frac{\pi}{18}})^6$

This gives:

${i\theta} = i\frac{\pi}{18}}*6$

${i\theta} = i\frac{6\pi}{18}}$

${i\theta} = i\frac{\pi}{3}}$

Divide through by i

$\theta = \frac{\pi}{3}$

Hence, the modulus, z is:

$|z| = 27$

And the argument $\theta$ is

$\theta = \frac{\pi}{3}$

4 0

3 years ago

If a and b are the measures of two first quadrant angles and sin a = 4/9 and sin b = 2/5 find sin(a+b).

frutty [35]

Answer:

$sin(a+b)=\frac{4\sqrt{21}+2\sqrt{65}}{45}$

Step-by-step explanation:

we are given

$sin(a)=\frac{4}{9}$

$sin(b)=\frac{2}{5}$

Firstly, we will draw triangle

we get

$cos(a)=\frac{\sqrt{65} }{9}$

$cos(b)=\frac{\sqrt{21} }{5}$

now, we can sue formula

$sin(a+b)=sin(a)cos(b)+cos(a)sin(b)$

now, we can plug values

$sin(a+b)=\frac{4}{9}\times \frac{\sqrt{21} }{5}+\frac{\sqrt{65} }{9}\times \frac{2}{5}$

now, we can simplify it

$sin(a+b)=\frac{4\sqrt{21}+2\sqrt{65}}{45}$

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irakobra [83]

Answer:

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