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3 years ago

Hello can you please help me posted picture of question

Mathematics

1 answer:

ankoles [38]3 years ago

8 0

Equation of the given line is:

y = 3x + 2

Slope of the line = m = 3

Since we are to find equation of line parallel to this line, the slope of new line will be the same as slope of parallel lines is the same.

So we are to find a line with slope 3 and passing through the point (10,1). Using the point-slope form of equation we can write:

y - 1 = 3(x - 10)
y = 3x - 30 + 1
y = 3x - 29
⇒
- 3x + y = 29

So option C gives the correct equation of the line.

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For the past 25 days, jonathon has read for n minutes each day. His total number of minutes read is 875. Write an equation to ex

liubo4ka [24]

Answer:

Required equation is: 25n = 875

Step-by-step explanation:

Let n be the number of minutes Jonathon read per day

Now

He read for 25 days and total 875 minutes

The equation will be:

$25n = 875$

We can solve this equation to find the number of minutes each day

Using Division property of Equality

$\frac{25n}{25} = \frac{875}{25}\\n = 35$

Hence,

Required equation is: 25n = 875

6 0

3 years ago

Can someone help me?

Tju [1.3M]

Answer:

Step-by-step explanation:

in table B if x = 1 y cannot equal both 1 and -1

no function can have x equal to two or more unique y's

7 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

Identify the correct trigonometry formula to use to solve for x.

Elan Coil [88]

Answer:

sin(62)=18/x

Step-by-step explanation: sine = opp/hyp

8 0

3 years ago

Read 2 more answers

In a parallelogram ABCD < D is a right angle. Is ABCD a rectangle?

Diano4ka-milaya [45]

Answer:

yes

Step-by-step explanation:

becoz D=90° as all side of rectangle have 90°

5 0

3 years ago