If vector u has its initial point at (-17, 4) and its terminal point at (8,-11), u = blank i+ blank j. If v = 14i-4j, 2(u - v) = blank i+ blank j

step-by-step explanation:

this is the formula: v = the value of investment at the end of the time period

p = the principal amount (the initial amount invested)

r = the annual interest rate

n = the annual frequency of compounding (how many times a year interest is added)

t = the number of years the money is invested

^ means raise to the power of

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.

sasasadaassasasasa opk ok you know