HELP PLSSS ITS URGENT!!!!!

Rewrite the function y= x^2 + 14x + 4 in vertex form

Step2247 [10]

Vertex form is basically commplete the square
y=a(x-h)^2+k

y=x^2+14x+4
take 1/2 of 14 and square it, (7^2=49)
add that and its negative to right side
y=x^2+14x+49-49+4
factor perfect squaer
y=(x+7)^2-49+4
y=(x+7)^2-45

answer is A

4 0

3 years ago

Name the property used to make the conclusion. If x+ 3 = 17, then x = 14 -

zimovet [89]

Answer:

Subtraction property of inequality

Step-by-step explanation:

Step 1: Definition of Subtraction property

Subtraction property of equality refers to balancing an equation by using the same mathematical operation (minus) on both sides.

Step 2: Relate the definition above with the given question.

It can be seen from the statements in the question that 3 was subtracted from both sides of the initial equation to get:

$x + 3 = 17 \\ x + 3 - 3 = 17 - 3 \\ x = 14$

6 0

3 years ago

Write a recursive rule for the sequence.

Vadim26 [7]

Answer:

f(n)=f(n-1)+f(n-2)

f(1)=1x

f(2)=1x

Step-by-step explanation:

This is the fibonacci sequence with each term times x.

Notice, you are adding the previous two terms to get the third term per consecutive triples of the sequence.

That is:

1x+1x=2x

1x+2x=3x

2x+3x=5x

3x+5x=8x

So since we need the two terms before the third per each consecutive triple in the sequence, our recursive definition must include two terms of the sequence. People normally go with the first two.

f(1)=1x since first term of f is 1x

f(2)=1x since second term of f is 1x

Yes, I'm naming the sequence f.

So I said a third term in a consecutive triple of the sequence is equal to the sum of it's two prior terms. Example, f(3)=f(2)+f(1) and f(4)=f(3)+f(2) and so on...

Note, the term before the nth term is the (n-1)th term and the term before the (n-1)th term is the (n-2)th term. Just like before the 15th term you have the (15-1)th term and before that one you have the (15-2)th term. That example simplified means before the 15th term you have the 14th and then the 13th.

So in general f(n)=f(n-1)+f(n-2).

So the full recursive definition is:

f(n)=f(n-1)+f(n-2)

f(1)=1x

f(2)=1x

8 0

3 years ago

Prove that if {x1x2.......xk}isany

Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where $n\geq 2$ . If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set $X= (x_1, x_2,....,x_n)$ is linearly dependent, so then by definition we have scalars $c_1,c_2,....,c_n$ in C such that:

$c_1 x_1 +c_2 x_2 +.....+c_n x_n =0$

And not all the scalars $c_1,c_2,....,c_n$ are equal to 0.

Since at least one constant is non zero we can assume for example that $c_1 \neq 0$ , and we have this:

$c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n$

We can divide by c1 since we assume that $c_1 \neq 0$ and we have this:

$v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n$

And as we can see the vector $v_1$ can be written a a linear combination of the remaining vectors $v_2,v_3,...,v_n$ . We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select $v_1$ and we have this:

$v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n$