Answer:
D. The rate of change for function B is greater than the rate of change for function A
Step-by-step explanation:
Function a has a slope (aka rate of change) of 3/3
I know this because we start at (-3, 0) then finish at (0, 3)
We added 3 to both the x value and the y value.
3/3 = 1, so we have a rate of change of 1 for function A
Now that we know this, we can write the equation out into slope intercept form.
We can get y = x + 3
The equation for function B is y = 5x + 5
The slope for function B is greater, therefore the rate of change for function B is greater than the rate of change for function A
Answer:
1024
Step-by-step explanation:
1 division: 2
2 divisions: 2×2 = 4
3 divisions: 2×(2×2) = 8
The number of viruses is 2^n, where n is the number of divisions.
After 10 divisions, there are 2^10 = 1024 viruses.
The distance is 14 between coordinates'
<span>y = slope*x + y-intercept;
</span>We can rewrite our equation in a shorter form : y = mx + b;
y = x + 2 ; m1 = 2 and b1 = 2;
y = -x + 6; m2 = -1 and b2 = 6;
<span>Set the two equations for y equal to each other:
</span>x + 2 = -x + 6 ;
<span>Solve for x. This will be the x-coordinate for the point of intersection:
</span>2x = 4;
x = 2;
<span>Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. This will be the y-coordinate of the point of intersection:
</span>y = 2 + 2 ;
y = 4;
<span>The point of intersection for these two lines is (2 , 4).</span>
Answer:
13
Step-by-step explanation:
where n is the number of terms, a1 is the first term and an is the last term. The sum of the first n terms of an arithmetic sequence is called an arithmetic series . Example 1: Find the sum of the first 20 terms of the arithmetic series if a1=5 and a20=62 .An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant. ... As with any recursive formula, the initial term of the sequence must be given. An explicit formula for an arithmetic sequence with common difference d is given by an=a1+d(n−1) a n = a 1 + d ( n − 1 ) .