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3 years ago

What is the midpoint between (7 − 3i) and ( 3 + 2i).

Mathematics

1 answer:

Rufina [12.5K]3 years ago

5 0

Answer:Answer: 5 - i/2

Step-by-step explanation:

Complex numbers on the complex plane are in essense the same, when it comes to midpoints. Applying the midpoint formula, we get the midpoint as ( 3+2i + 7-3i ) /2 , or (10-i)/2, or 5- i/2

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Simplify the expression.<br><br> can anyone explain to me how to do this? please

WINSTONCH [101]

Answer:

The answer is the last one (32x^7y^15)

You can bring x to the second power (x^2) because (x) is basically x^1. This is a basic exponent rule. (x^m)^n = x^m times n.

Then you can apply this rule to (2xy^3)^5. First you bring two to the fifth power and get 32. Then you bring x^5 according to the rule. Then you bring y^15, also because of the rule.

Now you have:

x^2 times 32x^5y^15

Now you just multiply the like terms together (x^2 and x^5)

When you multiply two exponents with the same base, you add the exponents together: a^n times a^m = a^n+m.

So you end up with 32x^7y^15

5 0

3 years ago

Read 2 more answers

I can’t figure out the surface area. Someone please help !! :)

Ilia_Sergeevich [38]

<h3> Answer:</h3><h2>surface area is 227•1</h2><h2>volume is 321•7</h2>

Step-by-step explanation:

diameter is 8•5

radius is d/2

8•5/2

=4•25

surface area = 4πr^2

4×22/7×4•25×4•25

=1589•5/7

=227•1

Volume= 4/3πr^3

4/3×22/7×4•25×4•25×4•25

4/3×22/7×76•76

=321•7

Hope it helps

plz mark as brainliest

5 0

2 years ago

A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = $\frac{AB}{BC} = \frac{h}{x}$

Taking inverse trigonometric function, α = tan⁻¹ ( $\frac{h}{x}$ ) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

$\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}$

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and $\frac{dh}{dt}$ = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

$\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512$