Answer:
15$
Step-by-step explanation:
35-10=25$
25-5=20$
20-5=15$
Answer:
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The maximum number of roots to a polynomial of order n is n roots. Take the example of a quadratic (order 2) which can intersect the x-axis a maximum of 2 times, and similarly a cubic (order 3) 3 times maximum.
Hence for 8 intersections, minimum order = 8
2 Simpify:
a -4 X x = -4x
b -10 X y = -10y
c -1 X a = -a
d b X (-1) = -b
e -4 X 2m = -8m
f 6 X -3a = -18a
g -8 X -3a = 24a
h -6m X 4 = -24m
i -7 X 8n = -56n
j -a X -3 = 3a
k 6x / -2 = -3x
l -10m / -5 = 2m
m -24a / 8 = -3a
n 2(m+3)-8=2(m)+2(3)-8=2m+6-8=2m-2
o 5(m-1)+9=5(m)+5(-1)+9=5m-5+9=5m+4
p 3(a-5)+10=3(a)+3(-5)+10=3a-15+10=3a-5
q 4(2x+1)-8x=4(2x)+4(1)-8x=8x+4-8x=4
r 3(10-2x)+3x=3(10)+3(-2x)+3x=30-6x+3x=30-3x
s 4(3-x)+9x=4(3)+4(-x)+9x=12-4x+9x=12+5x
3 Simplify by collecting like terms:
a 7a-5b+2a-6b=(7+2)a+(-5-6)b=(9)a+(-11)b=9a-11b
b 11x-2y-5x+7y=(11-5)x+(-2+7)y=(6)x+(5)y=6x+5y
c 3m+2g-5g-4m=(3-4)m+(2-5)g=(-1)m+(-3)g=-m-3g
d 6a-7-9a+10=(6-9)a+(-7+10)=(-3)a+(3)=-3a+3
e 7p-2q-6p+3q=(7-6)p+(-2+3)q=(1)p+(1)q=p+q
f 3x+7-12-5x=(3-5)x+(7-12)=(-2)x+(-5)=-2x-5
g 2ab+3bc-5ab+bc=(2-5)ab+(3+1)bc=(-3)ab+(4)bc=-3ab+4bc
h 6t^2+3t-5t^2-8t=(6-5)t^2+(3-8)t=(1)t^2+(-5)t=t^2-5t
i 9y-6z-9y+5z=(9-9)y+(-6+5)z=(0)y+(-1)z=0-z=-z
j 2k-3k^2-4k+k^2=(2-4)k+(-3+1)k^2=(-2)k+(-2)k^2=-2k-2k^2
k 10t+5w+t-7w=(10+1)t+(5-7)w=(11)t+(-2)w=11t-2w
l 7a-3b-8a-5b=(7-8)a+(-3-5)b=(-1)a+(-8)b=-a-8b
it is 40$$$Answer:
Step-by-step explanation:
