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bulgar [2K]
1 year ago
11

(2x-2) (x+5) what does x =

Mathematics
2 answers:
Oksana_A [137]1 year ago
7 0

Answer:

(2x-2)=

     +2 =+2

(2x)=2

----------

    2

x= 1

-----------------------------------------------

(x+5)=

   -5=-5

x= -5

Step-by-step explanation:

You just need to separate the variable from the numbers

Lynna [10]1 year ago
4 0

Answer:

a

x

2

+

b

x

+

c

=

0

the two roots of the equation take the form

x

1

,

2

=

−

b

±

√

b

2

−

4

a

c

2

a

So, start by adding

−

5

to both sides of the equation to get

2

x

2

+

x

−

5

=

5

−

5

2

x

2

+

x

−

5

=

0

Notice that you have

a

=

2

,

b

=

1

, and

c

=

−

5

. This means that the two solutions will be

x

1

,

2

=

−

1

±

√

1

2

−

4

⋅

2

⋅

(

−

5

)

2

⋅

2

x

1

,

2

=

−

1

±

√

41

4

You can simplify this if you want to get

x

1

=

−

1

+

√

41

4

≅

1.35078

and

x

2

=

−

1

−

√

41

4

≅

−

1.85078

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The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.
Mama L [17]

Answer:

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Step-by-step explanation:

Given;

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a + b + c = P

21 m  + 40 m  + c = 120 m

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                         B

                     ↓            ↓  

                  ↓                          ↓

                ↓                                       ↓

            A →  →  → →  →  → →  → →    →    →  C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0

Angle B is calculated as follows;

Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0

Angle C is calculated as follows;

Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0

Therefore, the largest angle of the field is 149⁰.

       

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