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4 years ago

Convert the polar representation of this complex number into its standard form:

Mathematics

1 answer:

7nadin3 [17]4 years ago

4 0

Answer:

Step-by-step explanation:

z = 2(cos7π/6 +isin7π/6)

cos7π/6 = cos(π/6+π) = - cos(π/6) = - √3 /2

sin7π/6 = sin(π/6+π) = - sin(π/6) = - 1 /2

so : z = 2( - √3 /2+i(- 1 /2))

z= - √3 - i .....(standard form)

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Gwar [14]

The answer is; 24/75 of the entire cake remained

If 2/5 of the initial cake was eaten, the 3/5 is what remained

MInu ate 1/3 of the 3/5 = 3/15

3/5 – 3/15 = 6/15 remained

Her brother ate; 1/5 of 6/15 = 6/75

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3 years ago

**The following scenario is for #12, #13, and #14 on this test.** A school chorus has 90 sixth grade students and 75 seventh gra

givi [52]

Answer:

The music director can make 15 groups of performers with 6 sixth grade students and 5 seventh grade students in each group.

Step-by-step explanation:

A school chorus has 90 sixth-grade students and 75 seventh-grade students.

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Find GCF(90,75):

Now,

Therefore, the music director can make 15 groups of performers with 6 sixth grade students and 5 seventh grade students in each group.

6 0

3 years ago

Think about how you can use absolute value notation to express the distance between two points on a coordinate graph. For each p

vagabundo [1.1K]

Answer:

(4, 9) and (-10, -8)

Step-by-step explanation:

8 0

3 years ago

7. The points P(2, -3), Q(3, -2) and R(8, z) are

Hatshy [7]

Answer:

z = 3

Step-by-step explanation:

Since the points are collinear then the slopes between the points are equal.

Calculate the slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = P (2, - 3) and (x₂, y₂ ) = Q (3, - 2)

m = $\frac{-2+3}{3-2}$ = 1

Repeat with

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3 years ago

Ignore circled one and work it out please

mash [69]

Answer:

a. $\frac{(x-7)}{x^2+x-30}$

Step-by-step explanation:

$\huge \frac{x^2-3x-28}{(x^2+10x+24)(x-5)} \\\\\huge {=\frac{x^2-7x+4x-28}{(x^2+6x+4x+24)(x-5)}}\\\\\huge {=\frac{x(x-7)+4(x-7)}{\{x(x+6)+4(x+6)\}(x-5)}}\\\\\huge {=\frac{(x-7)(x+4)}{(x+6)(x+4)(x-5)}}\\\\\huge {=\frac{(x-7)}{(x+6)(x-5)}}\\\\\huge {=\frac{(x-7)}{x^2+(6-5)x+6(-5)}}\\\\\huge {=\frac{(x-7)}{x^2+x-30}}$

3 0

3 years ago