I need help with this math question

Mathematics

1 answer:

givi [52]2 years ago

7 0

Answer:

a.) y = 3/(2ˣ)

b.) y = 1/(2ˣ)

c.) y = (π^π)ˣ

d.) y = (1/27)(1/√(3))ˣ

e.) y = .002908/(.119025ˣ)

f.) y = .00000004808/(.0413ˣ)

Step-by-step explanation:

Concept need to know is:

a negative exponent will flip the numerator with the denominator
a fraction as an exponent is just a root. so if the exponent is x^(1/2) then the root is 2 and x^(1/3) is a cube root

adding and subtracting exponent is the same thing as multiplying the same base.
so x^(1+2) = (x^1)(x^2)

and x^(1-2) = (x^1)(x^-2)

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Find the solutions to the equation below 30x^2-28x+6 =0

Naddik [55]

For this case, we have the following equation of the second degree:

$30x ^ 2-28x + 6 = 0$

If we divide between 2 on both sides of the equation, we will have:

$15x ^ 2-14x + 3 = 0$

Where:

$a = 15\\b = -14\\c = 3$

The solutions will come from:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Substituting:

$x = \frac {- (- 14) \pm \sqrt {(- 14) ^ 2-4 (15) (3)}} {2 (15)}\\x = \frac {14 \pm \sqrt {196-180}} {30}\\x = \frac {14 \pm \sqrt {16}} {30}\\x = \frac {14 \pm4} {30}$