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OLga [1]
3 years ago
10

The height of a triangle is 6 m more than its base. The area of the triangle is 56 m?. What is the length of the base? Enter you

r answer in the box.
Mathematics
1 answer:
Marizza181 [45]3 years ago
4 0

Answer:

Height = 14

Base = 8

Step-by-step explanation:

h=6+b

a=(1/2)bh

a=56

56=(1/2)bh

times 2 both sides

112=bh

h=6+b

112=b(6+b)

112=b²+6b

minus 112 both sides

0=b²+6b-112

factor

what 2 numbers multily to get -112 and

add to get 6

-8 and 14

0=(b-8)(b+14)

set to zero

b-8=0

b=8

b+14=0

b=-14

false, measures can't be negative

base=8

h=6+b

h=6+8

h=14

height=14

base=8

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These questions have asked us to solve by completing the square.
How do we? I have attached a picture, which will explain

6. x² + 2x = 8
→ b is the coefficient of x, which is 2
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x² + 2x + (\frac{2}{2} )^2 = 8 + ( \frac{2}{2})^2
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7. x² - 6x = 16

→ We do the same thing we did in the previous question

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8. x² - 18x = 19

x² - 18x + ( \frac{18}{2} )^2 = 19 + ( \frac{18}{2})^2
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\sqrt{(x-9)^2} =  \sqrt{100}
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9. x² + 3x = 3

x² + 3x + ( \frac{3}{2} )^2 = 3 +  (\frac{3}{2} )
x^2 + 3x +  \frac{9}{4} = 3 +  \frac{9}{4}
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(x^2 +  \frac{3}{2} ) ( x^2 +  \frac{3}{2} ) =  \frac{21}{4}
( x^2 + \frac{3}{2} )^2 =  \frac{21}{4}
\sqrt{ x^2 + \frac{3}{2} } =  \sqrt{ \frac{21}{4} } 
x +  \frac{3}{2} = + \frac{ \sqrt{21} }{2} or x +\frac{3}{2} = -  \frac{ \sqrt{21} }{2}
x = \frac{-3+ \sqrt{21} }{2} or x = \frac{-3 - \sqrt{21}}{2}


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Answer:

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Step-by-step explanation:

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when two lines cross like this they form vertical angles, and vertical angles are always equal.

calculate the system of equations and you'll get x=42 and y=23

To find the measure of any of the angles just put x or y into the equation.

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