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3 years ago

What is the area of the figure below

Mathematics

1 answer:

laila [671]3 years ago

5 0

Answer:

Hello! answer: area = 21

Step-by-step explanation:

5 × 3 = 15 3 × 2 = 6 15 + 6 = 21 HOPE THAT HELPS!

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Read 2 more answers

x + y = 42, xy = 108 then what is x and y. Is there any (non-quadratic)formula for finding this instead of the process of elimin

charle [14.2K]

Answer:

when x = 39.25, y = 2.75

when x = 2.75, x = 39.25

Step-by-step explanation:

use substitution method

x+y = 42 ------ (1)

xy = 108 ------ (2)

from (1) ,

x+y = 42

y = 42-x ------ (3)

substitute (3) into (2)

xy = 108

x( 42-x ) = 108

42x - x² = 108

-x² + 42x - 108 = 0

x² - 42x + 108 = 0

use the following formula to solve the value of x:

$i) \: x = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \: \\ ii) \: x = \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

a = 1

b = -42

c = 108

$i) \: x = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a}$

$x = \frac{ - ( - 42) + \sqrt{( - 42) {}^{2} - 4(1)(108)} }{2(1)}$

$x = \frac{42 + \sqrt{1764 - 432} }{2}$

$x = \frac{42 + \sqrt{1332} }{2}$

$x = \frac{42 + 6 \sqrt{37} }{2}$

$x = 21 + 3 \sqrt{37}$

$x = 39.25$

$ii) \: x \: \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

$x = \frac{ - ( - 42) - \sqrt{( - 42) {}^{2} - 4(1)(108} }{2(1)}$

$x = \frac{42 - \sqrt{1764 - 432} }{2}$

$x = \frac{42 - \sqrt{1332} }{2}$

$x = \frac{42 - 6 \sqrt{37} }{2}$

$x = 21 - 3 \sqrt{37}$

$x = 2.75$

the 2 values of x are x = 39.25 and x = 2.75

substitute x = 39.25 into (3)

y = 42-x

y = 42-39.25

y = 2.75

substitute x = 2.75 into (3)

y = 42-x

y = 42-2.75

y = 39.25

when x = 39.25, y = 2.75

when x = 2.75, x = 39.25

7 0

3 years ago

What is the area of the figure below​

What is the area of the figure below