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guajiro [1.7K]
3 years ago
14

true or false: the points (6, 13), and (21, 33) and (99, 137) all lie on the same line. th equation of the lines is y + /3x + 5.

explain or show your reasoning.
Mathematics
1 answer:
Phoenix [80]3 years ago
8 0

Answer:

point A (6,13) lies on the equation.  True

given point  B(21,33) lies on the equation.  True

given point  C (99, 137) lies on the equation.  True

The equation that was given was  y = 4/3x + 5

Now,check the given equation for the given points.

1)  A (6,13)

Substitute x = 6 in the given equation y = 4/3x +5

y = 4/3(6) + 5 = 4(2) + 5 = 8+5 = 13

y=13

the given point A (6,13) lies on the equation.

2)  B  (21,33)

Substitute x = 21 in the given equation y=4/3x+5

y=4/3(21) + 5 = 4(7) + 5=28+5=33

y = 33

the given point  B(21,33) lies on the equation.

3)  C (99, 137)

Substitute x = 99 in the given equation y = 4/3x + 5

y=4/3(99) + 5 = 4(33) + 5 = 132 + 5 = 137

y = 137

the given point  C (99, 137) lies on the equation.

There is your answer

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\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = \lim_{h\to0}\frac{(9+h)^4-9^4}h

Carry out the binomial expansion in the numerator:

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Then the 9⁴ terms cancel each other, so in the limit we have

\displaystyle \lim_{h\to0}\frac{4\times9^3h+6\times9^2h^2+4\times9h^3+h^4}h

Since <em>h</em> is approaching 0, that means <em>h</em> ≠ 0, so we can cancel the common factor of <em>h</em> in both numerator and denominator:

\displaystyle \lim_{h\to0}(4\times9^3+6\times9^2h+4\times9h^2+h^3)

Then when <em>h</em> converges to 0, each remaining term containing <em>h</em> goes to 0, leaving you with

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Alternatively, you can recognize the given limit as the derivative of <em>f(x)</em> at <em>x</em> = 9:

f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h \implies f'(9) = \lim_{h\to0}\frac{f(9+h)-f(9)}h

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