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densk [106]
3 years ago
5

How many square units are in the area of the triangle whose vertices are the x and y intercepts of the curve y = (x-3)^2 (x+2)?

Mathematics
1 answer:
pantera1 [17]3 years ago
6 0

The intercepts can be read from the equation as (-2, 0), (0, 18), (3, 0). These define a triangle with a base of 5 and an altitude of 18, so the area is

... A = (1/2)·5·18 = 45 . . . . square units


_____

The x-intercepts are those values of x that make one or another of the factors be zero. (x-3) is zero when x=3; (x+2) is zero when x=-2.

The y-intercept is the value when x=0. That is (-3)²·(2) = 9·2 = 18.

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Answer:

-50x

Step-by-step explanation:

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6 0
3 years ago
Determine whether or not 630 is a triangular number. start with the formula t n = n ( n + 1 ) 2 and then use the quadratic formu
ziro4ka [17]
Triangular sequence = n(n + 1)/2

If 630 is a triangular number, then:

n(n + 1)/2  = 630

Then n should be a positive whole number if 630 is a triangular number.

n(n + 1)/2  = 630

n(n + 1)  = 2*630

n(n + 1)  = 1260

n² + n = 1260

n² + n - 1260 = 0

By trial an error note that 1260 = 35 * 36

n² + n - 1260 = 0

Replace n with 36n - 35n

n² + 36n - 35n - 1260 = 0

n(n + 36) - 35(n + 36) = 0

(n + 36)(n - 35) = 0

n + 36 = 0   or   n - 35 = 0

n = 0 - 36,   or  n = 0 + 35

n = -36, or 35

n can not be negative. 

n = 35 is valid.

Since n is a positive whole number, that means 630 is a triangular number.

So the answer is True.
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Try adding a picture next time maybe...?

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Answer:

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Step-by-step explanation:

Since p(x) = x² + 2 , if x ≤ 4

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