The answer is a, all rhombuses have equal sides, if they have equal sides then day have equal angles.
Ok so this is conic sectuion
first group x's with x's and y's with y's
then complete the squra with x's and y's
2x^2-8x+2y^2+10y+2=0
2(x^2-4x)+2(y^2+5y)+2=0
take 1/2 of linear coeficient and square
-4/2=-2, (-2)^2=4
5/2=2.5, 2.5^2=6.25
add that and negative inside
2(x^2-4x+4-4)+2(y^2+5y+6.25-6.25)+2=0
factor perfect squares
2((x-2)^2-4)+2((y+2.5)^2-6.25)+2=0
distribute
2(x-2)^2-8+2(y+2.5)^2-12.5+2=0
2(x-2)^2+2(y+2.5)^2-18.5=0
add 18.5 both sides
2(x-2)^2+2(y+2.5)^2=18.5
divide both sides by 2
(x-2)^2+(y+2.5)^2=9.25
that is a circle center (2,-2.5) with radius √9.25
(a). To calculate Y at equilibrium
Y = C + I + G
Y = 40 + 0.8(Y – 0 + 10 + 20)
Y = 350
(b). To calculate C, I and G at Equilibrium
C = 40 + 0.8 Y
Since Y = 350
C = 40 + 0.8(350)
C = 40 + 280
C = 320
I = 20
G = 10
(c).To find
equilibrium Y
Given,
EX = 4 + 3EP/P
IM = 8 + 0.1 (Y - T) - 2EP/P
E = 3
P = 1
P = 1.5
Y = 170
First we find the common difference...to do this we subtract the first term from the second term. -7 - (-1) = -7 + 1 = -6
now we are going to find the 10th term
an = a1 + (n-1)*d <== formula for finding any term in arithmetic series
a1 = 1st term, d = common difference, n = term we want to find
now we sub
a10 = -1 + (10 -1) * -6
a10 = -1 + (9 * -6)
a10 = -1 - 54
a10 = - 55
now we will find the sum
Sn = (n (a1+ an)) / 2 <== formula for finding the sum
S10 = (10(-1 - 55))/2
S10 = (10(-56) / 2
S10 = -560/2
S10 = - 280
so the sum of the first 10 numbers is -280
Im confused could you explain what the questions are better.
