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Galina-37 [17]
3 years ago
8

Pls help me I will merl you as brain

Mathematics
1 answer:
Marizza181 [45]3 years ago
5 0

Answer:

it's the third one 99% Shure if not can you please tell me

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Number 6: Which is closer to 0.0009999? 0, 1/2, or 1?
aliya0001 [1]
0.
1/2 is equivalent to 0.5
1 is equivalent to 1.0

8 0
3 years ago
Which calculation results in the best estimate of 148 % of 203
adoni [48]
203---- 100%
x-----148%
x=203*148/100=203*1.48=300.44
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WILL MARK BRAINLIEST PLEASE HELP
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64-35= 29

so there were 29 boys included in the survey
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If (x^2 - 1) is a factor of ax^4 + bx^3 + cx^2 + dx + e, show that a + c + e = b + d = 0
alisha [4.7K]
x^2-1=x^2-1^2=(x-1)(x+1)

If x²-1 is a factor of the polynomial, both x-1 and x+1 are factors of it.

According to the remainder theorem, if a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

If x-1 and x+1 are factors of the polynomial p(x)=ax⁴+bx³+cx²+dx+e, then p(1)=0 and p(-1)=0.

p(1)=a \times 1^4+b \times 1^3 + c \times 1^2+ d \times 1+e \\
p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\
p(1)=a+b+c+d+e \\
p(-1)=a-b+c-d+e \\ \\
p(1)=0 \\
p(-1)=0 \\ \\ \hbox{add both equations:} \\
a+b+c+d+e=0 \\
\underline{a-b+c-d+e=0} \\
2a+2c+2e=0 \\
2(a+c+e)=0 \\
a+c+e=0 \\ \\
\hbox{substitute 0 for a+c+e in the first equation:} \\
a+b+c+d+e=0 \\
(a+c+e)+b+d=0 \\
0+b+d=0 \\
b+d=0 \\ \\
\boxed{a+c+e=b+d=0} \\
\hbox{proved } \checkmark
8 0
3 years ago
Pls help I already did G I just need help on J
olya-2409 [2.1K]

Answer: -3

-3(0)-1=-1

-3(-4)-1=11

-3(-6)-1=17

I hope this is good enough:

5 0
3 years ago
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