All categories

2 years ago

If x = (10 − 3i) and y = (3 − 10i), then xy= and xy=

Mathematics

2 answers:

Darina [25.2K]2 years ago

5 0

Answer:

$xy = -109i$

Step-by-step explanation:

Given that:

$x = 10 - 3i$

$y = 3-10i$

where, i is the imaginary part

We have to find xy.

$xy=(10-3i)(3-10i)$

⇒ $(10-3i)(3-10i) = 30-100i-9i+30i^2$

We know that: $i^2 = -1$

then;

⇒ $(10-3i)(3-10i) = 30-100i-9i-30=-109i$

⇒ $xy =-109i$

therefore, the product of x and y is, -109i

kipiarov [429]2 years ago

4 0

$xy=(10-3i)(3-10i)=30-100i-9i-30=-109i$

You might be interested in

We’re getting married bro

Montano1993 [528]

Answer:

It's not that hard so lets get a divorce bro

Step-by-step explanation:

7 0

2 years ago

QUESTION: Josephine put a frame round his picture which measured 36 cm by 34 cm. A margin 4cm wide was left all the way round. W

Mandarinka [93]

36x34= 1224cm²
(36-4)x(34-4)= 32x30= 960cm²

Area of frame= 1224
Area of picture inside frame= 960
So border = 1224-960= 264

ANSWER = 264cm²

5 0

2 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago

If a bus driver leaves her first stop by 7:00 a.m., her route will take less than 37 minutes. If she leaves after 7:00 a.m., she

lianna [129]

Answer:

i dont know lol

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Graph △ABC with vertices A(2, 3), B(−4, −3), and C(2, −3). Find the coordinates of the circumcenter.

Stolb23 [73]

Circumcenter = (-1,0)
The circumcenter of a triangle is the intersection of the perpendicular bisectors of the sides of the triangle. So let's calculate a couple of the bisectors and determine their intersection.
Slope AB = (3 - -3)/(2 - -4) = (3+3)/(2+4) = 6/6 = 1
Perpendicular bisector will have a slope of -1 and will pass through point ((2-4)/2,(3-3)/2) = -2/2,0/2) = (-1,0)
Equation is of the form
y = -x + b
Substitute known point
0 = -(-1) + b
0 = 1 + b
-1 = b
So the equation for the perpendicular bisector of AB is
y = -x - 1
Now let's calculate the perpendicular bisector of BC
Slope BC = (-3 - -3)/(-4 - 2) = (-3 + 3) / (-6) = 0/-6 = 0. This means that the
line is horizontal and that the perpendicular bisector will be a vertical line with infinite slope. A point that line will pass through is ((-4 + 2)/2, (-3 + -3)/2) =
(-2/2, 0/2) = (-1,0). So the equation for the line is:
x = -1
Now we want the intersection between
x = -1 and y = -x - 1
Since we know that x has to be -1, just substitute it into the 2nd equation.
y = -x - 1
y = -(-1) - 1
y = 1 - 1
y = 0

So the circumcenter is (-1,0).
Let's verify that. The distance from the circumcenter to each vertex of the triangle will be the same. Using the Pythagorean theorem, C^2 = A^2 + B^2. We're not going to bother taking the square root since if the squares are equal, then square roots will also be equal.
Distance^2 from (2,3):
(2- -1)^2 + (3-0)^2 = 3^2 + 3^2 = 9 + 9 = 18
Distance^2 from (-4,-3):
(-4 - -1)^2 + (-3 - 0)^2 = -3^2 + -3^2 = 9 + 9 = 18
Distance^2 from (2,-3):
(2 - -1)^2 + (-3 - 0)^2 = 3^2 + -3^2 = 9 + 9 = 18
The distances to all three vertexes is identical, so (-1,0) is verified as the circumcenter.

5 0

3 years ago