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oksian1 [2.3K]
3 years ago
10

Write an equation in slope-intercept form of the line having the given slope and y-intercept.

Mathematics
1 answer:
pychu [463]3 years ago
7 0

Answer:

The answer to your question is   y = 4/6x - 4

Step-by-step explanation:

Data

m = 4/6

P (0, -4)

slope-intercept form = ?      y = mx + b    where m is the slope and

                                                                  b is the y-intercept

Formula

           y - y1 = m(x - x1)

Substitution

           y - (-4) = 4/6(x - 0)

Simplification

           y + 4 = 4/6x

Result

                 y = 4/6x - 4

slope = 4/6  y-intercept = -4

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Step-by-step explanation:The formula to find the area of a triangle is Length times Base divided by 2. The length of the triangle could be 18 or 24, but that doesn’t matter. The base could also be 18 or 24, but that also doesn’t matter, because the hypotenuse (the longest part of a right triangle, in this case being 30), is not a part of the formula. 18 times 24 is 432, and 432 divided by 2 is 216. So the area is 216

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A series of three? separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers. Each
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Answer:

312.5\pi \text{ km}^3\approx 981.75\text{ km}^3

Step-by-step explanation:

We have been given that a series of 3 separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers.

Each of the three tunnels is shaped like a half-cylinder with a radius of 5 meters.

Since we know that volume of a semicircular or a half cylinder is half the volume of a circular cylinder.

\text{Volume of a semicircular cylinder}=\frac{\pi r^2h}{2}, where,

r = Radius of cylinder,

h = height of the cylinder.

Upon substituting our given values in volume formula we will get,

\text{Volume of a semicircular cylinder}=\frac{\pi (5\text{ km})^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*25\text{ km}^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*625\text{ km}^3}{2}

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

\text{Volume of a semicircular cylinder}=981.74770\text{ km}^3

Therefore, the volume of earth removed to build the three tunnels is 312.5\pi \text{ km}^3\approx 981.75\text{ km}^3.


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