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3 years ago

Solve for z over the imaginary numbers. Please help!

Mathematics

1 answer:

bezimeni [28]3 years ago

5 0

So you subtract the five over to get -5=2e^z, divide by 2, -2.5=e^z, take the natural log of both sides to get z=ln(-2.5) which does not exist

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How do i factor x^3+3x^2

Mariulka [41]

All you need to do is pull out the common factor of $x^2$ :
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4 0

3 years ago

Read 2 more answers

Help pls!! .........................................................................

ser-zykov [4K]

Answer:

a)8

b)13

Step-by-step explanation:

a) ar = c

a^2+b^2=c^2

4^2+7^2=c^2

16+49 = c2

c = $\sqrt{65}$ ~8

b) sa= c

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I hope im right!!

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3 years ago

Which of the following expressions represents the distance between 3/4 and 2

Anettt [7]

Answer:

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Not sure if this was what you were trying to ask but hope this helps

6 0

3 years ago

PLS SHOW UR SOLUTIONS

maksim [4K]

Answer:

Q1. $(2x-3)^{2}+3(x-4)=5+y\\$

$4x^{2} -12x+9+3x-12-5-y=0$

$4x^{2} -9x-8-y=0$

$4x^{2} -9x-8=y$

∴ $y = 4x^{2} -9x-8$

∴ From the quadratic equation above, a = 4, b = -9, and c = -8.

Q2. $(x-5)^{2} -x^{2} =y$

$x^{2} -10x+25-x^{2} =y$

$-10x+25=y$

∴ $y = -10x + 25$

∴ The equation above is not a quadratic equation as it does not follow the format of ax² + bx + c.

Q3. $2x(4x^{2} -4x+1)=y+x(8x)+4(2x)$

$8x^{3} -8x^{2} +2x-8x^{2} -8x=y$

∴ $y=8x^{3} -16x^{2} -6x$

∴ The equation above is not a quadratic equation as it does not follow the format of y = ax² + bx + c.

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1 year ago

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sergey [27]

100 = 15.50p + 7
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3 years ago