The third term in a sequence is 16.

Mathematics

1 answer:

REY [17]2 years ago

7 0

Answer:

s(n) = (5n) + 1

Step-by-step explanation:

s(n) = (5n) + 1 when n = 3 s(3) = 16

term to term is 5 more than the previous term "add 5"

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Does the function f(x) = 2x^2− 6x + 9 have a minimum or a maximum? Identify its value.

exis [7]

Answer:

minimum value of function is $\frac{45}{8}$ .

Step-by-step explanation:

Given function represents a parabola.

Now, here coefficient of $x^{2}$ is positive , so the parabola will be facing upwards and thus the function will be having a minimum.

Now, as we know that minimum value of a parabolic function occurs at

x = $\frac{-b}{4a}$ .

Where , b represents the coefficient of x and a represents the coefficient of $x^{2}$ .

here, a = 2 , b = -6

Thus $\frac{-b}{4a}$ = $\frac{6}{8}$

= $\frac{3}{4}$

So, at x = $\frac{3}{4}$ minimum value will occur and which equals

y = 2× $\frac{9}{16}$ - $\frac{18}{4}$ + 9 = $\frac{45}{8}$ .

Thus , minimum value of function is $\frac{45}{8}$ .

4 0

2 years ago

Use the quadratic formula to find the solutions to the quadratic equation below x^2-6x-5=0

Sonja [21]

Answer:

$\large\boxed{x=3-\sqrt{14}\ \vee\ x=3+\sqrt{14}}$

Step-by-step explanation:

$x^2-6x-5=0\qquad\text{add 5 to both sides}\\\\x^2-6x=5\\\\x^2-2(x)(3)=5\qquad\text{add}\ 3^2\ \text{to both sides}\\\\x^2-2(x)(3)+3^2=5+3^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-3)^2=5+9\\\\(x-3)^2=14\to x-3=\pm\sqrt{14}\qquad\text{add 3 to both sides}\\\\x=3-\sqrt{14}\ \vee\ x=3+\sqrt{14}$