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rewona [7]
2 years ago
15

Using quadratics, find the value of x. ​

Mathematics
2 answers:
Iteru [2.4K]2 years ago
4 0

Answer:

(x+1)+(3x+1)+(2x+4)=180 (sum od all sode of a triangle is 180)

x+1+3x+1+2x+4=180

6x+6=180

6x=180-6

x=174/6

x=66

Musya8 [376]2 years ago
3 0

Answer:

x = 4

Step-by-step explanation:

Using Pythagoras' identity on the right triangle.

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

(3x + 1)² = (x + 1)² + (2x + 4)² ← expand all factors using FOIL

9x² + 6x + 1 = x² + 2x + 1 + 4x² + 16x + 16 , simplify right side

9x² + 6x + 1 = 5x² + 18x + 17 ( subtract 5x² + 18x + 17 from both sides )

4x² - 12x - 16 = 0 ( divide through by 4 )

x² - 3x - 4 = 0 ← in standard form

(x - 4)(x + 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 4 = 0 ⇒ x = 4

x + 1 = 0 ⇒ x = - 1

However, x > 0 thus x = 4

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Lesechka [4]
C is the answer. if the water is being pumped out then the height of the water would decrease.
8 0
3 years ago
2/3 * 1/2+ 3/4 /1+ 1/6
In-s [12.5K]

Answer:

2/3 * 1/2+ 3/4 /1+ 1/6 =1.25

1.25

Step-by-step explanation:

3 0
2 years ago
Please help my teacher has been waiting for 30 full mniutes
anyanavicka [17]

Answer:

404.990 in^2

Step-by-step explanation:

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3 0
2 years ago
What is the constant rate of change of (5,80) (4,64) and (6,96)
sveticcg [70]

Answer:

16

Step-by-step explanation:

The existence of the constant rate of change is given the ratio of y to x is the same. Then:

k_{1} = \frac{80}{5}

k_{1} = 16

k_{2} = \frac{64}{4}

k_{2} = 16

k_{3} = \frac{96}{6}

k_{3} = 16

In consequence, the constant rate of change is 16.

8 0
2 years ago
Three forces of 300 N in the direction of N30E, 400N in the direction of N60E and
andreyandreev [35.5K]

Split up each force into horizontal and vertical components.

• 300 N at N30°E :

(300 N) (cos(30°) i + sin(30°) j)

• 400 N at N60°E :

(400 N) (cos(60°) i + sin(60°) j)

• 500 N at N80°E :

(500 N) (cos(80°) i + sin(80°) j)

The resultant force is the sum of these forces,

∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i

… … …  + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N

∑ F ≈ (546.632 i + 988.814 j) N

so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.

4 0
2 years ago
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