Which of the following is a conjugate for 7 + iSquare root of 2?

Mathematics

2 answers:

Mice21 [21]3 years ago

7 0

Answer:

7 - i $\sqrt{2}$

Step-by-step explanation:

Given a complex number a ± bi then the complex conjugate is

a ∓ bi

Note that the real part remains the same while the sign of the coefficient of the imaginary part is the opposite

Hence conjugate of 7 + i $\sqrt{2}$ is 7 - i $\sqrt{2}$

Orlov [11]3 years ago

4 0

7- i square root of 2

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The diagrams shows a 3cm x 5cm x 4cm cuboid.

Vera_Pavlovna [14]

Answer:

a) 5.83 cm

b) 34.45°

Step-by-step explanation:

a) From Pythagoras theorem of right triangles, given right triangle ABC:

AB² + BC² = AC²

Therefore:

AC² = 5² + 3²

AC² = 25 + 9 = 34

AC = √34

AC = 5.83 cm

b) From triangle ACD, AC = 5.83 cm, AD = 4 cm and ∠A = 90°.

From Pythagoras theorem of right triangles, given right triangle ACD:

AD² + AC² = DC²

Therefore:

DC² = 5.83² + 4²

DC² = 34 + 16 = 50

DC = √50

DC = 7.07 cm

Let ∠ACD be x. Therefore using sine rule:

$\frac{4}{sin(x)}=\frac{7.07}{sin(90)} \\ sin(x)=\frac{4*sin(90))}{7.07}=0.5657\\ x=sin^{-1}(0.5657)\\x=34.45^o$

6 0

2 years ago

Please answer this question correctly and i'll mark you as brainlilest! Thank you.

raketka [301]

Answer:

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