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ValentinkaMS [17]
3 years ago
10

There is a line that includes The point (1,7) and has a slope of 5 what is it’s equation in the slope-intercept form

Mathematics
1 answer:
Korolek [52]3 years ago
8 0

Answer:

y = 5x + 2

Step-by-step explanation:

The point-slope form of the equation of the line can be used to advantage. For slope m through point (h, k), that equation is ...

... y = m(x -h) +k

For your point and slope, this is ...

... y = 5(x -1) +7

Simplifying gives the desired equation.

... y = 5x +2

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Suppose y varies directly with x. If y = 6 when x = 2, find x when y = 12.
igomit [66]

Answer:

x = 4 when y = 12

Step-by-step explanation:

The ratio of y to x is 6:2. If you multiply either one of them, you must do the same to the remaining number. So, when you multiply 6 by 2 to get 12, you must also multiply 2 by 2 to keep the ratio equal and the same. 2x2 = 4, so x=4 when y=12. Hope this helped! Good luck with other math problems :)

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3 years ago
5.) - 7(-2x - 9) = 133<br> how do you solve this
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Answer: x= 5

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4 0
3 years ago
Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it
Zanzabum

Answer:

\beta=45\degree\:\:or\:\:\beta=135\degree

Step-by-step explanation:

We want to solve \tan \beta \sec \beta=\sqrt{2}, where 0\le \beta \le360\degree.

We rewrite in terms of sine and cosine.

\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}

\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}

Use the Pythagorean identity: \cos^2\beta=1-\sin^2\beta.

\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}

\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)

\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta

\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0

This is a quadratic equation in \sin \beta.

By the quadratic formula, we have:

\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }

\sin \beta=\frac{2}{2\cdot \sqrt{2} } or \sin \beta=\frac{-4}{2\cdot \sqrt{2} }

\sin \beta=\frac{1}{\sqrt{2} } or \sin \beta=-\frac{2}{\sqrt{2} }

\sin \beta=\frac{\sqrt{2}}{2} or \sin \beta=-\sqrt{2}

When \sin \beta=\frac{\sqrt{2}}{2} , \beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )

\implies \beta=45\degree\:\:or\:\:\beta=135\degree on the interval 0\le \beta \le360\degree.

When  \sin \beta=-\sqrt{2}, \beta is not defined because -1\le \sin \beta \le1

4 0
3 years ago
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