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Firdavs [7]
4 years ago
5

Please help immediately

Mathematics
1 answer:
Softa [21]4 years ago
5 0
Let a = arc distance of the sector
let r = radius of the circle


a + 2r = 8
a = (8-2r)

Find the portion of the circle in the sector (arc/circumference)

a/2r •r

Replace a with ( 8-2r)

(8-2r/2•square root • r)

Simply

(8-r/square root •r)

( 4-r) / square root R


Find the area of the sector
(4-r)/square root •r • square root •r ^2

Cancel pi*r


A = (4-r) * r
A = -r^2 + 4r

Max of this quadratic equation occurs at the axis of symmetry; x=-b/(2a)


R=-4/2•(-1)

R= 2 cm

Find the max area

A = -2^2 + 8(2)
A = -4+ 16
A = 12 sq cm max area
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Without looking it up- what is -2317 - (-5294)?
vovangra [49]

Answer:

=2977

Step-by-step explanation:

-2317 - (-5294)

-2317 + 5294

2977

4 0
3 years ago
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What fraction of the first twenty counting numbers (1,2,3,...,18,19,20) are composite numbers​
drek231 [11]

Answer:

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6 0
3 years ago
Find the slope through the points (2,4) and (6,12)
jeyben [28]

Answer:

The slope is 2

Step-by-step explanation:

Slope = \frac{rise(changeiny)}{run(changeinx)} =\frac{y2-y1}{x2-x1} =\frac{12-4}{6-2} =\frac{8}{4}

8/4 reduces to 2/1 which makes the slope 2

8 0
3 years ago
Lines k and p are perpendicular, neither is vertical and p passes through the origin. Which is greater? A. The product of the sl
AlladinOne [14]
What do we know about those two lines?

They are perpendicular, meaning they have the same slope.
We know the slope of both is not zero (neither is vertical).
Therefore either
1) Both slopes are positive and therefore the product is positive
2) Both slopes are negative and therefore the product is positive (minus by a minus is a plus)

For the y intercepts, we know that the line P passes through the origin.
Therefore its Y intercept is zero.
[draw it if this is not obvious and ask where does it cross the y axis]
Therefore the Y intercept of line K and line P is zero.
[anything multiplied by a zero is a zero]

So we know that the product of slopes is positive, and we know that the product of Y intercepts is zero.
So the product of slopes must be greater.
Answer A
4 0
3 years ago
Which expression is a cube root of -1+i√3?
Tpy6a [65]

Answer:

<em>The correct option is C.</em>

Step-by-step explanation:

<u>Root Of Complex Numbers</u>

If a complex number is expressed in polar form as

Z=(r,\theta)

Then the cubic roots of Z are

\displaystyle Z_1=\left(\sqrt[3]{r},\frac{\theta}{3}\right)

\displaystyle Z_2=\left(\sqrt[3]{r},\frac{\theta}{3}+120^o\right)

\displaystyle Z_3=\left(\sqrt[3]{r},\frac{\theta}{3}+240^o\right)

We are given the complex number in rectangular components

Z=-1+i\sqrt{3}

Converting to polar form

r=\sqrt{(-1)^2+(\sqrt{3})^2}=2

\displaystyle tan\theta=\frac{\sqrt{3}}{-1}=-\sqrt{3}

It's located in the second quadrant, so

\theta=120^o

The number if polar form is

Z=(2,120^o)

Its cubic roots are

\displaystyle Z_1=\left(\sqrt[3]{2},\frac{120^o}{3}\right)=\left(\sqrt[3]{2},40^o\right)

\displaystyle Z_2=\left(\sqrt[3]{2},40^o+120^o\right)=\left(\sqrt[3]{2},160^o\right)

\displaystyle Z_3=\left(\sqrt[3]{2},40^o+240^o\right)=\left(\sqrt[3]{2},280^o\right)

Converting the first solution to rectangular coordinates

z_1=\sqrt[3]{2}(\ cos40^o+i\ sin40^o)

The correct option is C.

8 0
4 years ago
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