Answer:
Ok, first in our series we can see two numbers in the Sigma, one bellow 0, and other above, 4.
This means that the value of k will go from 0 to 4, then all the numbers in the sum are:
(-1/2)^0 + (-1/2)^1 + (-1/2)^2 + (-1/2)^3 + (-1/2)^4
So we have 5 terms in our series.
b) to see the sign in each term, we must solve the powers, remember that:
(-1)^n is -1 if n is odd, and is equal to 1 if n is even, so we have:
(-1/2)^0 + (-1/2)^1 + (-1/2)^2 + (-1/2)^3 + (-1/2)^4
= 1 -1/2 + 1/4 - 1/8 + 1/16.
So the sign in each term of the series alternates.
First to get the equation you knew to understand one thing about perpendicular lines. The slope of the line is the opposite reciprocal of the perpendicular lines or the new slope is m = 10.
Then you use the formula
y = mx + b
you plug in your values from the point and the new slope.
(1,5) with new slope m
5= 10(1)+b
5-10=b
-5 = b
then make your new equation
y = 10x -5
that's your line that goes through point (1,5) and is perpendicular to the line given
Answer:
It is expected that linearization beyond age 20 will be use a function whose slope is monotonously decreasing.
Step-by-step explanation:
The linearization of the data by first order polynomials may be reasonable for the set of values of age between ages from 5 to 15 years, but it is inadequate beyond, since the fourth point, located at 
, in growing at a lower slope. It is expected that function will be monotonously decreasing and we need to use models alternative to first order polynomials as either second order polynomic models or exponential models.
Answer:
Rosaria purchased 50 bracelets and 70 necklaces
Step-by-step explanation:
Let the number of bracelets be b and the number of necklaces be n
b + n = 120 •••••(i)
Secondly;
10b + 11n = 1270 ••••(ii)
Total cost of b bracelets at 10 per 1 is 10b
Total cost of n bracelets at 11 per 1 is 11n
Adding both gives 1270
From i, b = 120-n
Substitute this into ii
10(120-n) + 11n = 1270
1200 - 10n + 11n = 1270
n = 1270-1200
n = 70
b = 120-n
b = 120-70
b = 50
