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3 years ago

What is (4-3i)(2+I)\4+3i

Mathematics

1 answer:

Olegator [25]3 years ago

6 0

(4-3i)(2+l)/4+3i > (8-6i+4l-3il)/4+3i > (8+6i+4l-3il)/4 > 2+3i/2+l-3il final answer

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What do you think you need to do to the

8090 [49]

Answer:

You need to multiply the terms together, therefore adding the exponents 2 and 3 to get 5.

Step-by-step explanation:

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3 years ago

Can someone help me with section b please and thank you

blondinia [14]

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3 years ago

Try the numbers 22, 23, 24, 25 in the equation 4/3 = 32/d to test whether any of them is a solution. a. 23 is a solution. c. 24

S_A_V [24]

Answer:

c. 24 is a solution

Step-by-step explanation:

Given equation:

$\frac{4}{3}=\frac{32}{d}$

To test the numbers 22, 23, 24, 25 for the solution.

Solution:

In order to test the given number for the solution, we will plugin each number in the unknown variable $d$ and see if it satisfies the equation.

1) $d=22$

$\frac{4}{3}=\frac{32}{22}$

Reducing fraction to simplest form by dividing the numerator and denominator by their G.C.F.

$\frac{4}{3}=\frac{32\div 2}{22\div 2}$

$\frac{4}{3}=\frac{16}{11}$

The above statement can never be true and hence 22 is not a solution.

2) $d=23$

$\frac{4}{3}=\frac{32}{23}$

The fractions can no further be reduced.

The statement can never be true and hence 23 is not a solution.

3) $d=24$

$\frac{4}{3}=\frac{32}{24}$

Reducing fraction to simplest form by dividing the numerator and denominator by their G.C.F.

$\frac{4}{3}=\frac{32\div 8}{24\div 8}$

$\frac{4}{3}=\frac{4}{3}$

The above statement is always true and hence 24 is a solution.

4) $d=25$

$\frac{4}{3}=\frac{32}{25}$

The fractions can no further be reduced.

The statement can never be true and hence 25 is not a solution.

4 0

3 years ago

Given:

lisov135 [29]

Answer:

Step-by-step explanation:

6 0

3 years ago

(ab^2 x -b^4a^3)^2<br><br> What is the answers

Umnica [9.8K]

Answer:

$\left(ab^2x-b^4a^3\right)^2:\quad a^2b^4x^2-2a^4b^6x+a^6b^8$

Step-by-step explanation:

Given the expression

$\left(ab^2\:x\:-b^4a^3\right)^2$

solving the expression

$\left(ab^2\:x\:-b^4a^3\right)^2$

$\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a-b\right)^2=a^2-2ab+b^2$

$a=ab^2x,\:\:b=b^4a^3$

so the expression becomes

$=\left(ab^2x\right)^2-2ab^2xb^4a^3+\left(b^4a^3\right)^2$

$=a^2b^4x^2-2a^4b^6x+a^6b^8$

Thus,

$\left(ab^2x-b^4a^3\right)^2:\quad a^2b^4x^2-2a^4b^6x+a^6b^8$

3 0

3 years ago