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mestny [16]
2 years ago
5

Give the equation of the line in slope intercept form. y = 3x + [?] =​

Mathematics
1 answer:
Ray Of Light [21]2 years ago
6 0

Answer:

See below:

Step-by-step explanation:

Hello! I hope you are having a nice day. My name is Galaxy and I will be helping you today.

We can solve this problem in a single step, we just need to know how equations like this are set up.

Equations in slope intercept form are set up in y=mx+b form. With m being the slope of the line, and b being the y intercept.

In your problem, we need to find the value of b, which is the y intercept.

The y intercept is what is y when x is zero. We can find that by looking at the graph supplied.

When we look at the line, we can see that it hits the x axis when y is negative two, therefore the intercept is (0,-2).

Since we know that, we can reconstruct our equation with this information, we already know the slope so we can add in -2 to get our answer.

y=3x-2

This would be our final answer.

Cheers!

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A rectangle has a height of 5x and a width of x + 2. Express the area of the entire rectangle.
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8 0
2 years ago
You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b
Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

\mathbf{x = 100\ m}

\mathbf{\frac{dy}{dt} = 2\ ms^{-1}} ---- the rate

\mathbf{\theta = \frac{\pi}{4}}

First, we calculate the vertical distance (y) using tangent ratio

\mathbf{\tan(\theta) = \frac{y}{x}}

Substitute 100 for x

\mathbf{y = 100\tan(\theta) }

\mathbf{\tan(\theta) = \frac{y}{100}}

Differentiate both sides with respect to time (t)

\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}

Substitute values for the rates and \mathbf{\theta }

\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}

This gives

\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}

\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}

Divide both sides by 2

\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }

\mathbf{ \frac{d\theta}{dt} = 0.01 }

Hence, the rate of change between you and the ball is 0.01 rad per second

Read more about related rates at:

brainly.com/question/16981791

8 0
2 years ago
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