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Vladimir79 [104]
3 years ago
8

The sum of two numbers is 21 . One number is 12 more than the other one. Find the numbers.

Mathematics
2 answers:
vaieri [72.5K]3 years ago
8 0
The 2 numbers would be 9 and 12
labwork [276]3 years ago
7 0

Answer:

4.5 and 16.5

Step-by-step explanation:

4 + 16 = 20

5 + 17 = 22

(NUMBERS HAVE TO BE BETWEEN 4 - 5 AND 16 - 17)

4.5 + 16.5 = 21

16.5 - 12 = 4.5

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Vladimir79 [104]

Answer:

If solving for x then x = -3/2 +y/2

If solving for y then y = -3-2x

Step-by-step explanation:

For x: Divide both sides by 2

For y: Subtract 2x from both sides of the equation

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Lines intersected by a transversal can never be parallel lines. True or false
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Answer : True

When lines intersect they can't be parallel.!

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Find all solutions and solutions in interval[0,2pi)
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7 0
3 years ago
For the function f(x) = -2x?, if the domain is {-3, 0, 3}, find the range.<br> =-
Nutka1998 [239]

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4 0
3 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}

6 0
3 years ago
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