Answer: $29.70
Step-by-step explanation:
250 times the annual interest (12%) is converted to 250 times .12 = 29.70
You just have to plug in and do it you can search it online!
The answer would be A. When using Cramer's Rule to solve a system of equations, if the determinant of the coefficient matrix equals zero and neither numerator determinant is zero, then the system has infinite solutions. It would be hard finding this answer when we use the Cramer's Rule so instead we use the Gauss Elimination. Considering the equations:
x + y = 3 and <span>2x + 2y = 6
Determinant of the equations are </span>
<span>| 1 1 | </span>
<span>| 2 2 | = 0
</span>
the numerator determinants would be
<span>| 3 1 | . .| 1 3 | </span>
<span>| 6 2 | = | 2 6 | = 0.
Executing Gauss Elimination, any two numbers, whose sum is 3, would satisfy the given system. F</span>or instance (3, 0), <span>(2, 1) and (4, -1). Therefore, it would have infinitely many solutions. </span>
Answer:
Point Critical point
Q (2,0) local minimum
R (-2,1) Saddle
S (2,-1) local maximum
T ( -2,-1) Saddle
O ( -2,0) Saddle
Step-by-step explanation: INCOMPLETE ANSWER INFORMATION ABOUT THE POINTS ARE RARE
f(x,y) = x³ +y⁴ - 6x -2y² +3
df/dx = f´(x) = 3x² -6x
df/dxdx = f´´(xx) = 6x
df/dy = f´(y) = -4y
df/dydy = 4
df/dydx = df/dxdy = 0
df/dydy = f´´(yy)
D = [ df/dxdx *df/dydy] - [df/dydx]²
D = (6x)*4 - 0
D = 6*2*4 D > 0 and the second derivative on x is 6*2 = 12
so D > 0 and df/dxdx >0 there is a local minimum in P
Q(2,1)
D = (6*2)*4 D>0 and second derivative on x is 6*2
The same condition there is a minimum in Q
R ( -2,1)
D = 6*(-2)*4 = -48 D< 0 there is a saddle point in R
S (2,-1)
D = 6*2*4 = 48 D > 0 and df/dxdx = 6*-1 = -6
There is a maximum in S
T ( -2,-1)
D = 6*(-2)*(4) = -48 D<0 there is a saddle point in T
O ( -2,0)
D = 6*(-2)*4 = -48 D<0 there is a saddle point in O
Answer:she would have at least gotten a 75 to pass.
Step-by-step explanation:76+82+71+75=304
304÷4= a solid 76%
