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3 years ago

What is the common ratio in this geometric sequence? 3,12,48

Mathematics

1 answer:

Katen [24]3 years ago

4 0

It has a 1:4 ratio I believe

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3. In a lake, there is a patch of lily pads. Every day, the patch doubles in size. If it takes 48 days for the patch to cover th

SSSSS [86.1K]

The patch doubles in size each day so, on the 47th will be half the size than on the 48th day

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3 years ago

<img src="https://tex.z-dn.net/?f=x%20%7B%7D%5E%7B2%7D%20%20-%202ax%20%2B%202%28a%20%7B%7D%5E%7B2%7D%20%20-%206%29%20%3D%200" id

pentagon [3]

answer

$- {4a}^{2} + 48$

5 0

3 years ago

Help please. and thank you

alexandr402 [8]

Answer:

Its Option C

Step-by-step explanation:

Basically Linear is derived from the word line

Also this an equation for the line

The y-intercept is 8 and the gradient is -7

This is why the function is linear

5 0

2 years ago

Read 2 more answers

How many solutions does x2 - 10x=-25 have?

Elina [12.6K]

Answer:

4.2 Solving x2-10x+25 = 0 by Completing The Square . This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.

Step-by-step explanation:

3 0

2 years ago

Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

3 0

3 years ago

Read 2 more answers