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Norma-Jean [14]
3 years ago
8

Compare and contrast between x-intercept and y-intercept

Mathematics
2 answers:
NemiM [27]3 years ago
6 0
The intercepts of a graph are points at which the graph crosses the axes. The x-intercept is the point at which the graph crosses the x-axis. ... The y-intercept is the point at which the graph crosses the y-axis. At this point, the x-coordinate is zero.
slamgirl [31]3 years ago
4 0

Answer:

"The x-intercept is the point at which the graph crosses the x-axis... ...The y-intercept is the point at which the graph crosses the y-axis." (Lumen)

The x-intercept is on the horizontal axis while the y-intercept is vertical axis. Aside from that they are practically the same and serve the same purpose while graphing, they just go in differing positions.

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Note that both 12 and 30 are multiples of 2. Divide the numerator and denominator by 2. 6/15. Note that 6 and 15 are multiples of 3. Divide the numerator and denominator by 3. 2/5. Note that 2 and 5 are both prime numbers. Your answer is 2/5.
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For what side length(s) is the area of an equilateral triangle equal to 30 cm?? Only enter the number, in centimeters, rounded t
xenn [34]

Answer: The sides length are 8.32 cm

Step-by-step explanation:

An equilateral triangle has all his sides of the same lenght, so we assume that the triangle has an L lenght in his sides.

The area of a triangle is Area = \frac{base * height}{2} where the base is L, the Area is 30 and an unknown height.

To determine the height, we cut the triangle in half and take one side. By simetry, one side has a base of \frac{L}{2}, a hypotenuse of L and a the unknown height.  

Then we apply the <em>Pythagoras theorem</em>, this states that <em>in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides</em>, or, hypotenuse = \sqrt{c^{2} + c^{2} } Where one c is \frac{L}{2} and the other is the height.

Then we find one of the c of the equation wich will be the height.

height = \sqrt{hypotenuse^{2}-base^{2} }

height = \sqrt{ L^{2} -\frac{L}{4} ^{2}}\\height = \sqrt{\frac{ 3L^{2}}{4} } \\\\height = \frac{\sqrt{3}L }{2}

Finally, we use the triangle area mentioned before an find the value of L.

30 = \frac{L*\frac{\sqrt{3}L }{2} }{2} \\\\L = \sqrt{\frac{120}{\sqrt{3} } } \\\\L = 8.32 cm

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