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kotykmax [81]
2 years ago
12

Please help me with this

Mathematics
2 answers:
SashulF [63]2 years ago
8 0

Answer:

512

Step-by-step explanation:

2x2x2x2x2x2x2x2x2= 512

sleet_krkn [62]2 years ago
8 0
512! :) have a great day!
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PLEASE HELP WILL GIVE BRAINLIIEST 20 POINTS 6TH GRADE ASAP PLEASE
nignag [31]

Answer:

Both A and B

Step-by-step explanation:

For functions you are just trying to make sure all the x's are different.  

So let's look at A:  The x's there are all different so it is a function

Now let's look at B: The x's there are all different so it is a function.

Both A and B

6 0
3 years ago
I need help please :)
padilas [110]

Answer:

what the heck is that hard math what grade are you even in?

Step-by-step explanation:

3 0
3 years ago
Please help? I'm. terrible at these
kykrilka [37]
13.5 = 30% of 45
4.5 = 15% of 30
4.6 = 23% of 20
4.2 = 60% of 7

Hope this helps!!
3 0
3 years ago
CAN SOMEONE PLEASE HELP ME???<br> ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
OLEGan [10]

Answer:

Option D, AP

Step-by-step explanation:

<u>The two sides that are connected by the right angle are called the sides.  The longest one that is not connect with a right angle is called the hypotenuse.</u>

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In this case, the sides are AZ and PZ.  <em>The hypotenuse is AP</em>

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Answer:  Option D, AP

6 0
3 years ago
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 \\&#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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