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mrs_skeptik [129]
3 years ago
9

Heelppppppppp!!!!! Please?!

Mathematics
2 answers:
finlep [7]3 years ago
6 0

To calculate the area shaded green, we first calculate the sum of areas S=A1+A2+A3.   The green area is the area of circle A less the sum of white areas S.

Also, note that the

area of an equilateral triangle with side s

=(sqrt(3)*s)*s/2

=sqrt(3)*s^2/2.

Proceeding,

A1

=sqrt(3)s^2/2

=sqrt(3)7^2/2

=49sqrt(3)/2

A2

=60/360 of the area of the circle

=60/360*(pi r^2)

=1/6*(pi*7^2)

=49pi/6

A3=A1

=49sqrt(3)/2

S=A1+A2+A3

=49sqrt(3)/2+49pi/6+49sqrt(3)/2

=49sqrt(3)+49pi/6

=49(sqrt(3)+pi/6)

Area shaded green

= pi r^2 - S

=pi 7^2 - S

=49pi - 49(sqrt(3)+pi/6)

=49(5pi/6-sqrt(3))


Gre4nikov [31]3 years ago
3 0

To calculate h evaluate inside the square root before taking the square root

h = √(7² - 3.5²) = √(49 - 12.25) = √36.75 = 6.06 ( to 2 dec. places)


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51

Step-by-step explanation:

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After reading 80% of her e-mails in her inbox, Danette still has M unread emails. Which of the following expressions could repre
Fofino [41]

Answer:

m/(1 - .8)

Step-by-step explanation:

She read 80% of her emails, which is .8 of the total.  So her unread emails would be 100% - 80% = 1 - .8

that means me can be written as:

m = (1 - .8)t

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If we solve for t, we get:

t = m/(1 - .8)

4 0
3 years ago
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Write the standard form of an equation of an ellipse subject to the given conditions.
irina [24]

Answer:

The answer is below

Step-by-step explanation:

The standard form of the equation of an ellipse with major axis on the y axis is given as:

\frac{(x-h)^2}{b^2} +\frac{(y-k)^2}{a^2} =1

Where (h, k) is the center of the ellipse, (h, k ± a) is the major axis, (h ± b, k) is the minor axis, (h, k ± c) is the foci and c² = a² - b²

Since the minor axis is at (37,0) and (-37,0), hence k = 0, h = 0 and b = 37

Also, the foci is at (0,5) and (0, -5), therefore c = 5

Using c² = a² - b²:

5² = a² - 37²

a² = 37² + 5² = 1369 + 25

a² = 1394

Therefore the equation of the ellipse is:

\frac{x^2}{1369}+ \frac{y^2}{1394} =1

6 0
3 years ago
PLEASE HELP!! URGENT!! i will mark brainliest if its right!! In the figure below, ∠DEC ≅ ∠DCE, ∠B ≅ ∠F, and DF ≅ BD. Point C is
zvonat [6]

Answer:

See below.

Step-by-step explanation:

This is how you prove it.

<B and <F are given as congruent.

This is 1 pair of congruent angles for triangles ABC and GFE.

<DEC and <DCE are given as congruent.

Using vertical angles and substitution of transitivity of congruence of angles, show that angles ACB and GEF are congruent.

This is 1 pair of congruent angles for triangles ABC and GFE.

Now you need another side to do either AAS or ASA.

Look at triangle DCE. Using the fact that angles DEC and DCE are congruent, opposite sides are congruent, so segments DC and DE are congruent. You are told segments DF and BD are congruent. Using segment addition postulate and substitution, show that segments CB and EF are congruent.

Now you have 1 pair of included sides congruent ABC and GFE.

Now using ASA, you prove triangles ABC and GFE congruent.

7 0
3 years ago
Help me please and thank you!
lorasvet [3.4K]

Answer:

-4/3

Step-by-step explanation:

-2 - 1 is -3

6-2 is 4

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