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3 years ago

Square root -64 minus square root -25... complex numners.. i am so confused

Mathematics

2 answers:

allochka39001 [22]3 years ago

7 0

The answer would be 3

Galina-37 [17]3 years ago

3 0

$\sqrt{ - 64} - \sqrt{ - 25} \\ 8i - 5i \\ 3i \\$
square root of negative is an imaginary solution or number. meaning it exist somewhere. square root of 64 is 8 and the square root of 25 is 5. "i" represent imaginary number

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What is each power of i with its multiplicative inverse.

pentagon [3]

Answer:

1) multiplicative inverse of i = -i

2) Multiplicative inverse of i^2 = -1

3) Multiplicative inverse of i^3 = i

4) Multiplicative inverse of i^4 = 1

Step-by-step explanation:

We have to find multiplicative inverse of each of the following.

1) i

The multiplicative inverse is 1/i

if i is in the denominator we find their conjugate

$=1/i * i/i\\=i/i^2\\=We\,\, know\,\, that\,\, i^2 = -1\\=i/(-1)\\= -i$

So, multiplicative inverse of i = -i

2) i^2

The multiplicative inverse is 1/i^2

We know that i^2 = -1

1/-1 = -1

so, Multiplicative inverse of i^2 = -1

3) i^3

The multiplicative inverse is 1/i^3

We know that i^2 = -1

and i^3 = i.i^2

$1/i^3\\=1/i.i^2 \\=1/i(-1)\\=-1/i * i/i\\=-i/i^2\\= -i/-1\\= i$

so, Multiplicative inverse of i^3 = i

4) i^4

The multiplicative inverse is 1/i^4

We know that i^2 = -1

and i^4 = i^2.i^2

$=1/i^2.i^2\\=1/(-1)(-1)\\=1/1\\=1$

so, Multiplicative inverse of i^4 = 1

3 0

3 years ago

Factor the trinomial <br> 5x2+11x+6

pentagon [3]

Answer:

factor by grouping

(x-1) (5x-6)

Step-by-step explanation:

correct me if i am wrong

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3 years ago

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miskamm [114]

Answer:

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Step-by-step explanation:

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3 years ago

Simplify the equation ( 3(3) × 3 (4) ) ÷ 3 (2)

aksik [14]

Answer:

(3(3) × 3(4) ) ÷ 3(2)

(9 × 12) ÷ 6

108 ÷ 6

3 0

2 years ago

Read 2 more answers

The two non-parallel sides of an isosceles trapezoid are each 7 feet long. The longer of the two bases measures 22 feet long.

katen-ka-za [31]

Answer:

Let X represent the shorter base, and drop the heights from shorter to longer base.

Two congruent right triangles are formed, because the trapezoid is isosceles and

the heights are the same.

The base angles are then 70 degree each, as they total 140.

So the height is 7* sin 70 = 6.5778483455....

The base of each right triangle is 11 - x/2

Now the algebra gets ugly.

By pythagorean theorem:

(11 - x/2)^2 + (7sin70)^2 = 7^2

(11-x/2)^2 + 49sin70^2 = 49

(11 - x/2)^2 = 49 - 49sin70^2

(11 - x/2)^2 = 49(1 - sin70^2)

(11 - x/2)^2 = 49 cos70^2 <-- trig identity : sin^2 + cos^2 = 1 ---> cos^2 = 1 - sin^2

121 - 11x + x^2/4 = 49 cos70^2

484 - 44x + x^2 = 196 cos 70^2

x^2 - 44x + 484 - 196cos70^2 = 0

x^2 - 44x + C = 0 where C = 484 - 196cos70^2 = 461.07235

By quadratic formula:

X = 44 +OR- square-root( (-44)^2 - 4*C) / 2

= 44 +OR- square-root( 91.71)/2

The two roots are:

26.78282 AND 17.2117331

The first one is disqualified and we must reject it

because if used will make the base of the

right triangles negative. Remember this is

supposed to be the smaller base. So it

cannot be greater than 22.

SO the smaller base is 17.2117331

So 11 - x/2 = overhang = 2.39413345

Drawing the diagonal, it's length can

be found by pythagorean theorem.

square-root(17.2117331)^2 + (7sin70)^2)=18.4258472....<-- diagonal

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3 years ago