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Fofino [41]
4 years ago
5

What is the equation if x=1 and y=16/3

Mathematics
1 answer:
GaryK [48]4 years ago
4 0
There may be more than one way in which to answer this question.  I will assume that the "equation" is a linear one:  f(x) = mx + b.

Then (16/3) = m(1) + b
This is one equation in two unknowns, so it does not have a unique solution.  Was there more to this problem than you have shared?


If we assume that the y-intercept (b) is zero, then y = mx, and 

16/3 = 1m, so that m = 16/3, and so   y = (16/3)x.
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How many real solutions does the equation 8x^2 − 10x + 15 = 0 have?
Nimfa-mama [501]

Answer:

The correct option is (A) No real solution.

Step-by-step explanation:

The expression provided is a quadratic equation.

8x^{2}-10x+15=0

The roots of a quadratic equation are:

x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }

Here,

a = 8

b = -10

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The conditions to determine real and complex roots are:

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Compute the value of b^{2}-4ac as follows:

b^{2}-4ac=(-10)^{2}-(4\times 8\times15)\\\\

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The equation has two complex roots.

Thus, the correct option is (A).

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