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3 years ago

7

Write an equation for the line that is parallel to the given line and passes through the given point

2 answers:

Lana71 [14]3 years ago

8 0

Y = 5x+4. The slope is the same, 5, making them parallel, and when you plug in 2 for x you get 14

victus00 [196]3 years ago

3 0

Answer:

$y = 5x+4$

Step-by-step explanation:

For this case we have the following equation:

$y = 5x +10$

And we want a line parallel to this one that passes thourgh the point (2,14). In order to satisfy this we need that condition:

$m_1 = m_2 =5$

Where $m_1 = 5$ on this case for the original slope, and if we solve for $m_2$ we got:

$m_2 = 5$

And now since we have the new slope we can solve for the value of b like this:

$14 = 5*(2) + b$

And if we solve for b we got:

$b = 14- 10= 4$

And then our equation parallel is given by:

$y = 5x+4$

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if you do this and explain how to do it i will rait you 5 stars and give you a brainlyieist it has to be simplify 2(5+3x)+(x+10)

Paha777 [63]

Answer:

20+7x

Step-by-step explanation:

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3 years ago

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a_sh-v [17]

Answer:

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Step-by-step explanation:

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3 years ago

Simplify the radical expression √(5) + 6√(5)

Evgesh-ka [11]

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4 years ago

Find 6% of $5.65 and round the product to the nearest cent.

Leviafan [203]

Answer:

$0.34

Step-by-step explanation:

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8 0

4 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

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

Where:

$\tan(\theta) = 9$ ---- given

So:

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

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago