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Mazyrski [523]
3 years ago
7

20 points!!!

Mathematics
2 answers:
Ghella [55]3 years ago
8 0
Area of semi circle: you know the radius is 5. 5^2 * pi = 25pi/2 = 12.5pi
For triangle: bh/2 = (10 * 10)/2 = 50
Add them together: 12.5pi + 50
12.5pi = 39.25
39.25 + 50 = 89.25
It’s closest to D
Svetach [21]3 years ago
8 0

Area of equilateral triangle is =

\sqrt{3}  \div 4 \times s {}^{2}

Root 3/4*10*10

Root 3/4*100

Root 3*25

25 root 3

Area of semicircle =1/2*22/7*5²

=1/2*22/7*25

(11*25)/7

275/7

=39.28

Adding areas of triangle and semicircle

25 root 3 +39.28

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A parallelogram has base (2x - 1) metres and height (4x - 7) metres.
Jobisdone [24]

Answer:

0.4069 ; 1.843

Step-by-step explanation:

Given:

Base of parallelogram, b = (2x - 1)

Height = (4x - 7)

Area = 1

Area of parallelogram = Base * height

Area of parallelogram = (2x - 1) * (4x - 7)

(2x - 1) * (4x - 7) = 1

8x² - 14x - 4x + 7 = 1

8x² - 18x + 7 - 1 = 0

8x² - 18x + 6 = 0

Divide through by 2

4x² - 9x + 3 = 0

Solving the quadratic equation :

Using the formula

-b ± √(b² - 4ac) / 2a

a = 4 ; b = - 9 ; c = 3

Plugging in the values :

-(-9) ± √((-9)² - 4(4)(3)) / 2(4)

9 ± √(81 - 48) / 8

9 ± √33 / 8

(9 ± 5.7445626) / 8

(9 - 5.7445626) / 8) = 0.4069

(9 + 5.7445626) / 8 = 1.843

4 0
3 years ago
PRECALC QUESTION! Convert the polar representation of this complex number into its standard form: z = 2(cos 11pi/6+ i sin 11pi/6
Tanzania [10]
\bf z=2\left[ cos\left( \frac{11\pi }{6} \right) + i\ sin\left( \frac{11\pi }{6} \right) \right]\qquad 
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\theta=\frac{11\pi }{6}\\
-----\\
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b=y=rsin(\theta)
\end{cases}\implies a+bi
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\boxed{\sqrt{3}-i}
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