All categories

2 years ago

ASAP WILL GIVE BRAINLIEST

Mathematics

2 answers:

professor190 [17]2 years ago

5 0

Answer:

(1,3)

Step-by-step explanation:

It is just where the points fall on the graph

natka813 [3]2 years ago

5 0

Answer:

isn't it (3,1) then since it's (x,y)

Step-by-step explanation:

You might be interested in

What's the distance between (0,2) and (0,8)?

N76 [4]

Answer:

6 units

Step-by-step explanation:

The distance between two points can be calculated with the formula

$\boxed{\textcolor{steelblue}{\text{Distance between 2 points}= \sqrt{(y_1 - y_2)^{2} + (x_1 - x_2)^{2} } }}$

However, since (0, 2) and (0, 8) have the same x coordinate of 0, the distance between them is the difference in their y-coordinates.

Distance between (0, 2) and (0, 8)

= 8 -2

= 6 units

7 0

2 years ago

What is 6x+78/67-78x=180+270-567

Sonja [21]

Answer:

15.31

Step-by-step explanation:

solve both sides separately

6x+78/67-78x

x cancles out

becomes 84/-11

becomes -7.64 rounded

180+270-567

becomes -117

moves -7.64 to other side (divide by other side)

-117/-7.64

becomes 15.31

8 0

2 years ago

Convert the polar representation of this complex number into its standard form z=4(cos150+isin150).

Katen [24]

Answer:

Step-by-step explanation:

Using the exact values

cos150° = - cos30° = - $\frac{\sqrt{3} }{2}$

sin150° = sin30° = $\frac{1}{2}$

Given

z = 4(cos150° + isin150° ) , substitute values

= 4(- $\frac{\sqrt{3} }{2}$ + $\frac{1}{2}$ i )

= - 2 $\sqrt{3}$ + 2i , that is

(- 2 $\sqrt{3}$ , 2 ) → B

7 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$