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Vinil7 [7]
3 years ago
13

Write the function f(x)=x^2−4x−7 in vertex form

Mathematics
1 answer:
Whitepunk [10]3 years ago
6 0

Answer:

y=(x - 2) squared - 11

Step-by-step explanation:

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Geometry question, need some help :)
kodGreya [7K]

Answer:

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Step-by-step explanation:


8 0
4 years ago
Which equation is equivalent to n + 4 = 11? *
dangina [55]

Answer:

(n + 4) *2 = 11 * 2

Step-by-step explanation:

(n + 4) *2 = 11 * 2

Divide both sides by 2

\frac{(n + 4) *2}{2} = \frac{11 * 2}{2}

n + 4 = 11

5 0
3 years ago
Transversal cuts parallel lines and at points X and Y respectively. The point nearest to C is X, and the point nearest to D is Y
hammer [34]

Answer:  The answer is (B) ∠SYD.


Step-by-step explanation: As mentioned in the question, two parallel lines PQ and RS are drawn in the attached figure. The transversal CD cut the lines PQ and RS at the points X and Y respectively.

We are given four angles, out of which one should be chosen which is congruent to ∠CXP.

The angles lying on opposite sides of the transversal and outside the two parallel lines are called alternate exterior angles.

For example, in the figure attached, ∠CXP, ∠SYD and ∠CXQ, ∠RYD are pairs of alternate exterior angles.

Now, the theorem of alternate exterior angles states that if the two lines are parallel having a transversal, then alternate exterior angles are congruent to each other.

Thus, we have

∠CXP ≅ ∠SYD.

So, option (B) is correct.

5 0
3 years ago
Please answer below if you know it! I am desperate
sveta [45]
A line of symmetry basically is like a reflection of a shape, so your answer is D.
5 0
3 years ago
Use mathematical induction to show that 4^n ≡ 3n+1 (mod 9) for all n equal to or greater than 0
cestrela7 [59]
When n=0, you have

4^0=1\equiv3(0)+1=1\mod9

Now assume this is true for n=k, i.e.

4^k\equiv3k+1\mod9

and under this hypothesis show that it's also true for n=k+1. You have

4^k\equiv3k+1\mod9
4\equiv4\mod9
\implies 4\times4^k\equiv4(3k+1)\mod9
\implies 4^{k+1}\equiv12k+4\mod9

In other words, there exists M such that

4^{k+1}=9M+12k+4

Rewriting, you have

4^{k+1}=9M+9k+3k+4
4^{k+1}=9(M+k)+3k+3+1
4^{k+1}=9(M+k)+3(k+1)+1

and this is equivalent to 3(k+1)+1 modulo 9, as desired.

3 0
4 years ago
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