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timofeeve [1]
3 years ago
13

I need help on 8 and 9 plz help me

Mathematics
1 answer:
pochemuha3 years ago
3 0
7 is just adding.
8 is counting by odd numbers.
9 is a rotation
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If 5(2x - 1) = 35, then x =​
SIZIF [17.4K]

Answer:

x = 4

Step-by-step explanation:

Given

5(2x - 1) = 35 ( divide both sides by 5 )

2x - 1 = 7 ( add 1 to both sides )

2x = 8 ( divide both sides by 2 )

x = 4

5 0
3 years ago
What is the domain of the function?<br><br> y= sqrt 5x-10
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I don't know if this will help but I found this on YAHOO!

Answer: Domain of definition of a function is the set of numbers which the variable attains and for which the function is defined.

Step-by-step explanation:

f(x) = sqrt (5x-10)

Here x can have value equal to any real number >=2 because if x attains value less that 2, 5x-10 becomes negative and sqrt(5x-10) has no real value.

therefore the domain of the function f(x) is (2, infinity) inclusive of 2.

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Help me find the equation! I did the work first and submitted it early, the teacher gave me the answer I was looking for but I h
sergeinik [125]

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A=1/2 bh

Step-by-step explanation:

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3 years ago
What is the average speed of a boat that travels 57 kilometers in 3 hours?
Blababa [14]

Answer:

19 km/h

Step-by-step explanation:

average speed = total distance / total time

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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
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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 \\&#10;p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\&#10;p(1)=a+b+c+d+e \\&#10;p(-1)=a-b+c-d+e \\ \\&#10;p(1)=0 \\&#10;p(-1)=0 \\ \\ \hbox{add both equations:} \\&#10;a+b+c+d+e=0 \\&#10;\underline{a-b+c-d+e=0} \\&#10;2a+2c+2e=0 \\&#10;2(a+c+e)=0 \\&#10;a+c+e=0 \\ \\&#10;\hbox{substitute 0 for a+c+e in the first equation:} \\&#10;a+b+c+d+e=0 \\&#10;(a+c+e)+b+d=0 \\&#10;0+b+d=0 \\&#10;b+d=0 \\ \\&#10;\boxed{a+c+e=b+d=0} \\&#10;\hbox{proved } \checkmark
8 0
3 years ago
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