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Tems11 [23]
3 years ago
14

How do you solve - 5d + 8e =-21 3d - 4e = 15 with system of equations

Mathematics
2 answers:
Mama L [17]3 years ago
8 0

Answer:

d = 9

e = 3

Step-by-step explanation:

- 5d + 8e =-21

  3d - 4e = 15  | ×2

________________ +

- 5d + 6d + 8e - 8e = - 21 + 30

d = 9

3d - 4e = 15

3×9 - 4e = 15

4e = 27 - 15

4e = 12

e = 3

satela [25.4K]3 years ago
6 0

Answer:

d = 9, e = 3

Step-by-step explanation:

Given the 2 equations

- 5d + 8e = - 21 → (1)

3d - 4e = 15 → (2)

Multiplying (2) by 2 and adding to (1) will eliminate the term in e

6d - 8e = 30 → (3)

Add (1) and (3) term by term to eliminate e, that is

d = 9

Substitute d = 9 into either of the 2 equations and solve for e

Substituting into (2)

3(9) - 4e = 15

27 - 4e = 15 ( subtract 27 from both sides )

- 4e = - 12 ( divide both sides by - 4 )

e = 3

Thus d = 9 and e = 3

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11

Step-by-step explanation:

3 0
3 years ago
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Fittoniya [83]

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3 years ago
Two investments totallng $47,500 produce an annual income of $3000. One investment ylelds 9% per year, while the other yields 6%
enot [183]

Answer:

The correct answer is "$5000".

Step-by-step explanation:

The given values are:

Two investments totaling,

= $47,500

Annual income,

= $3000

One investment yields per year,

= 9%

Other yield,

= 6%

Let,

  • The amount invested in 9% will be "x".
  • The amount invested in 6% will be "47,500-x".

Now,

⇒  0.09x+0.06(47,500-x)=3000

⇒        0.09x+2850-0.06x=3000

⇒                     0.03x+2850=3000

⇒                                0.03x=150

⇒                                       x=\frac{150}{0.03}

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5 0
2 years ago
Calculate two iterations of Newton's Method for the function using the given initial guess. (Round your answers to four decimal
pickupchik [31]

Answer:

The first and second iteration of Newton's Method are 3 and \frac{11}{6}.

Step-by-step explanation:

The Newton's Method is a multi-step numerical method for continuous diffentiable function of the form f(x) = 0 based on the following formula:

x_{i+1} = x_{i} -\frac{f(x_{i})}{f'(x_{i})}

Where:

x_{i} - i-th Approximation, dimensionless.

x_{i+1} - (i+1)-th Approximation, dimensionless.

f(x_{i}) - Function evaluated at i-th Approximation, dimensionless.

f'(x_{i}) - First derivative evaluated at (i+1)-th Approximation, dimensionless.

Let be f(x) = x^{2}-8 and f'(x) = 2\cdot x, the resultant expression is:

x_{i+1} = x_{i} -\frac{x_{i}^{2}-8}{2\cdot x_{i}}

First iteration: (x_{1} = 2)

x_{2} = 2-\frac{2^{2}-8}{2\cdot (2)}

x_{2} = 2 + \frac{4}{4}

x_{2} = 3

Second iteration: (x_{2} = 3)

x_{3} = 3-\frac{3^{2}-8}{2\cdot (3)}

x_{3} = 2 - \frac{1}{6}

x_{3} = \frac{11}{6}

7 0
3 years ago
Paolo is using his grandmother’s cookie recipe. He always doubles the amount of chocolate chips and oats. The recipe calls for 2
lesya692 [45]

Answer:

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Step-by-step explanation:

Hope this helps.

6 0
3 years ago
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