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LUCKY_DIMON [66]
3 years ago
15

Solve the equation 14=7(2x-4) using two different methods. Show your work. Which method do you prefer? Explain. need help thanks

Mathematics
2 answers:
Jet001 [13]3 years ago
5 0

Answer:

x=3

Step-by-step explanation:

14=7(2x-4)

14/7=2x-4

2=2x-4

2x=2+4

2x=6

x=6/2

x=3

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

14=7(2x-4)

14=14x-28

14x=14+28

14x=42

x=42/14

x=3

vodka [1.7K]3 years ago
3 0

14=7(2x-4)\\2=2x-4\\2x=6\\x=3\\\\14=7(2x-4)\\14=14x-28\\14x=42\\x=3

I prefer the first method, because I don't do unnecessary multiplication and therefore operate on smaller numbers.

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Bella cycled a total of 12 kilometers by making 4 trips to work. How many trips will Bella have to make to cycle a total of 18 k
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6 trips because 1 trip = 3 km
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6 0
3 years ago
A number,M, is rounded to 1 decimal place. the result is 9.4 complete the error interval for M​
ivann1987 [24]

Answer:

9.35 \le M \le 9.45

Step-by-step explanation:

Given

M = 9.4 to 1 d p

Required

The error interval

The error interval is represented as:

Lower \le M \le Upper

Where

Lower = M - 0.05

Upper = M + 0.05

So, we have:

Lower = M - 0.05

Lower = 9.4 - 0.05

Lower = 9.35

Upper = M + 0.05

Upper = 9.4 + 0.05

Upper = 9.45

Hence, the error interval: Lower \le M \le Upper is:

9.35 \le M \le 9.45

3 0
2 years ago
Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes
azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

  • The centre of the ellipse is at the origin and the X axis is the major axis

  • It passes through the points (-3, 1) and (2, -2)

<h2>To Find:</h2>

  • The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1

Given that the ellipse passes through the point (-3, 1)

Hence,

\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1

Cross multiplying we get,

  • 9b² + a² = 1 ²× a²b²
  • a²b² = 9b² + a²

Multiply by 4 on both sides,

  • 4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1

Cross multiply,

  • 4b² + 4a² = 1 × a²b²
  • a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

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Substituting in 2,

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  • 32/3 b² = 4b² + 128/3
  • 32/3 b² = (12b² + 128)/3
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  • 20b² = 128
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Substituting the values in the equation for ellipse,

\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1

\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
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