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3 years ago

−5x+4y=38 −x+2y=10 solve by elimination

Mathematics

1 answer:

tatyana61 [14]3 years ago

8 0

Answer:

(-6, 2)

Step-by-step explanation:

-5x + 4y - 38 = 0 (1) | -x +2y - 10 = 0 (2)

First i multiply the 2nd equation by -5, because i want the magnitude of the x coefficient to be the same but with opposite signs, because when i'll add the equations the x's are going to cancel.

(2) * (-5):

5x - 10y + 50 = 0 (2')

(1) + (2'):

-5x + 4y - 38 + 5x - 10y + 50 = 0

-6y + 12 = 0

y = 2

Use the second equation, before multiplying to find the x - solution.

-x + 2(2) - 10 = 0

-x + 4 - 10 = 0 | add 'x' to both sides

x = -6

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Step-by-step explanation:

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3 years ago

Carolina biked 1 mile more than twice the number of miles James biked. Carolina biked a total of 5 miles. Write an equation to d

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2 years ago

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

nydimaria [60]

Answer:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

Step-by-step explanation:

Given

$\cos(\theta) = -\frac{2}{3}$

$\theta \to$ Quadrant III

Required

Determine $\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

We have:

$\cos(\theta) = -\frac{2}{3}$

We know that:

$\sin^2(\theta) + \cos^2(\theta) = 1$

This gives:

$\sin^2(\theta) + (-\frac{2}{3})^2 = 1$

$\sin^2(\theta) + (\frac{4}{9}) = 1$

Collect like terms

$\sin^2(\theta) = 1 - \frac{4}{9}$

Take LCM and solve

$\sin^2(\theta) = \frac{9 -4}{9}$

$\sin^2(\theta) = \frac{5}{9}$

Take the square roots of both sides

$\sin(\theta) = \±\frac{\sqrt 5}{3}$

Sin is negative in quadrant III. So:

$\sin(\theta) = -\frac{\sqrt 5}{3}$

Calculate $\csc(\theta)$

$\csc(\theta) = \frac{1}{\sin(\theta)}$

We have: $\sin(\theta) = -\frac{\sqrt 5}{3}$

So:

$\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}$

$\csc(\theta) = \frac{-3}{\sqrt 5}$

Rationalize

$\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}$

$\csc(\theta) = \frac{-3\sqrt 5}{5}$

So, we have:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)$

Substitute: $\csc(\theta) = \frac{-3\sqrt 5}{5}$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}$

Take LCM

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

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abruzzese [7]

Answer:

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Step-by-step explanation:

As you know the compound interest formula is ...

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$2210.44 -1522.16 = $688.28 . . . . . . card I increased more

The difference in the increases is ...

$688.28 -605.78 = $82.50

Card I increased by $82.50 more than card H in the time period.

5 0

3 years ago