The n-th term of a number pattern is as follows:

Mathematics

1 answer:

Ilia_Sergeevich [38]2 years ago

7 0

9 = 3² is not prime, so $T_9 = 9^2-8 = 81-8 = 73$ .

7 is prime, so $T_7 = 7\times7+6 = 55$ .

Then the difference is $T_9 - T_7 = 73 - 55 = \boxed{18}$ .

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A biker moves 2.5 times faster than a walker. They started moving simultaneously in the same direction from the same point. In t

saveliy_v [14]

Answer:

The speed of the biker is 15

Step-by-step explanation:

First, you call the speed of the walker x, and the speed of the biker 2.5x.

Then you know that speed times time equals distance, so you set up and equation. You do x times 2, which is 2x, and then 2.5x times 2, which is 5x. Then, since the distance between them is 18 miles, the equation would be 5x-2x=18. You would get 3x=18, and x is 6. So 6 is the speed of the walker, and 6 times 2.5 = 15, so the speed of the biker is 15.

5 0

3 years ago

Read 2 more answers

Solve for x in the equation x squared + 2 x + 1 = 17.

Papessa [141]

Answer:

$x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\$

Step-by-step explanation:

given equation

$x^2 +2x +1 = 17$

subtracting 17 from both sides

$x^2 +2x +1 = 17\\x^2 +2x +1 -17= 17-17\\x^2 +2x - 16 = 0\\$

the solution for quadratic equation

$ax^2 + bx + c = 0$ is given by

x = $x = -b + \sqrt{b^2 - 4ac} /2a \\\\and \ \\-b - \sqrt{b^2 - 4ac} /2a$

________________________________

in our problem

a = 1

b = 2

c = -16

$x =( -2 + \sqrt{2^2 - 4*1*-16}) /2*1 \\x =( -2 + \sqrt{4 + 64}) /2\\x =( -2 + \sqrt{68} )/2\\x = ( -2 + \sqrt{4*17} )/2\\x = ( -2 + 2\sqrt{17} )/2\\x = - 1 + \sqrt{17}\\and\\\\x = - 1 - \sqrt{17}\\$