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3 years ago

Please Help Can someone explain where the 4i to the square root of 3 came from?

Mathematics

1 answer:

Y_Kistochka [10]3 years ago

5 0

Answer:

see explanation

Step-by-step explanation:

note that $\sqrt{-1}$ = i

$\sqrt{64-112}$

= $\sqrt{-48}$

= $\sqrt{16(3)(-1)}$

= $\sqrt{16}$ × $\sqrt{3}$ × $\sqrt{-1}$

= 4 × $\sqrt{3}$ × i

= 4i $\sqrt{3}$

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3 years ago

2 3/5 as a decimal.

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Answer:

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3 years ago

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Given 3(x-2)=6, then x is equal to: Select one: a. 5 b. 4 c. 2 d. 6

GenaCL600 [577]

3.(x - 2) = 6

3.x - 3.2 = 6

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4 years ago

A circle has a radius of \sqrt{13}√ 13 square root of, 13, end square root units and is centered at (-9.3,4.1)(−9.3,4.1)le

timofeeve [1]

Answer: The equation of the circle is

$x^2+y^2+18.6x-8.2y+90.3=0.$

Step-by-step explanation: We are given to write the equation of the circle with radius √13 units and center at the point (-9.3, 4.1).

We know that

the standard equation of a circle with radius r units and center at the point (h, k) is given by

$(x-h)^2+(y-k)^2=r^2.$

In the given circle,

radius, r = √13 units and center, (h, k) = (-9.3, 4.1).

Therefore, the equation of the circle will be

$(x-(-9.3))^2+(y-4.1)^2=(\sqrt{13})^2\\\\\Rightarrow (x+9.3)^2+(y-4.1)^2=13\\\\\Rightarrow x^2+18.6y+86.49+y^2-8.2y+16.81=13\\\\\Rightarrow x^2+y^2+18.6x-8.2y+103.3=13\\\\\Rightarrow x^2+y^2+18.6x-8.2y+90.3=0.$

Thus, the equation of the circle is

$x^2+y^2+18.6x-8.2y+90.3=0.$

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3 years ago

121+b^2=12.1^2<br> What is the value of b^2?

luda_lava [24]

$121 + b^2 = 12.1^2$
Square 12.1, and subtract 121 from both sides.
$b^2 = 146.41 - 121$
Complete the subtraction on the right side.
$b^2 = 25.41$
This is the solution.

$b^2 = 25.41$

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4 years ago