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sertanlavr [38]
3 years ago
11

Quadratic fucntion is represented by y=4(4x+1)^2+3. what is the equivalent form of this fuction.

Mathematics
1 answer:
In-s [12.5K]3 years ago
7 0

Answer:

64x^2+32x+7

Step-by-step explanation:

(4x+1)2= (4x+1)(4x+1)=16x^2+8x+1

4(16x^2+8x+1)=3

64x^2+32x+4+3

64x^2+32x+7

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I can't figure out this answer, can someone help?
raketka [301]
To solve this problem, we need to know that the adjacent interior angles (2 and 5,  or 3 and 4) between two parallel lines are supplementary.
This means that ∠2+∠5 = 180 °

which means 
x+50 + 4x+30 = 180
collect similar terms and solve
5x + 80 = 180
5x = 180-80 = 100
x=20 °

∠2 = x+50 = 20+50 = 70 °

But ∠ 1 = ∠ 2  (vertical angles)
therefore
∠ 1 = 70 °

8 0
3 years ago
Solve -24.8 + a = 49.67.
AleksandrR [38]

Answer:

A= 74.47

Step-by-step explanation:

you add the 24.8 to the right side to get a by itself

7 0
3 years ago
Read 2 more answers
Z^m-n (z^m+ z^m+n + z^n)
rusak2 [61]

Answer:Your left hand side evaluates to:

m+(−1)mn+(−1)m+(−1)mnp

and your right hand side evaluates to:

m+(−1)mn+(−1)m+np

After eliminating the common terms:

m+(−1)mn from both sides, we are left with showing:

(−1)m+(−1)mnp=(−1)m+np

If p=0, both sides are clearly equal, so assume p≠0, and we can (by cancellation) simply prove:

(−1)(−1)mn=(−1)n.

It should be clear that if m is even, we have equality (both sides are (−1)n), so we are down to the case where m is odd. In this case:

(−1)(−1)mn=(−1)−n=1(−1)n

Multiplying both sides by (−1)n then yields:

1=(−1)2n=[(−1)n]2 which is always true, no matter what n is

8 0
3 years ago
Find all the zeros of the polynomial function p(x) = x3 – 5x2 + 33x – 29
hram777 [196]

Answer:

\large \boxed{\sf \ \ x=1, \ \ x=2+5i, \ \ x=2-5i \ \ }

Step-by-step explanation:

Hello,

I assume that we are working in \mathbb{C}, otherwise there is only one zero which is 1. Please consider the following.

First of all, <u>we can notice that 1 is a trivial solution</u> as

   p(1) = 1^3-5\cdot 1^2 + 33\cdot 1-29=1-5+33-29=0

It means that (x-1) is a factor of p(x) so we can find two real numbers, a and b, so that we can write the following.

p(x)=(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=x^3+(a-1)x^2+(b-a)x-b

Let's identify like terms as below.

a-1 = -5 <=> a = -5 + 1 = -4

b-a = 33

-b = -29 <=> b = 29

So

\boxed{ \ p(x)=(x-1)(x^2-4x+29) \ }

Now, we need to find the zeroes of the second factor, meaning finding x so that:

x^2-4x+29=0 \ \text{ complete the square, 29 = 25 + 4}  \\ \\  x^2-2\cdot 2 \cdot x+2^2+25=0 \\ \\ (x-2)^2=-25=(5i)^2 \ \text{ take the root } \\ \\x-2=\pm 5i \ \text{ add 2 } \\ \\  x = 2+5i \ \text{ or } \ x = 2-5i

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

5 0
3 years ago
15 points<br>Steps needed
morpeh [17]
It is not a negative so it is 10 because the number can't be big
3 0
3 years ago
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