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notka56 [123]
3 years ago
10

Find the vertex and zeros of the following function f(x)=x^2-6x-7

Mathematics
1 answer:
elena-s [515]3 years ago
6 0

Try this solution:

Step-by-step explanation:

1) for zeros of the function: x²-6x-7=0; ⇒ x₁=7; x₂= -1.

2) for the vertex of the function: x₀= -b/2a, where a;b;c - numbers from ax²+bx+c=0 (from x²-6x-7=0).

x₀=6/2=3;

y₀= (4ac-b²)/4a; ⇒ y₀= -16.

It means, the coordinates of the vertex are (3;-16).

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Find the first, fourth, and eighth terms of the sequence. A(n) = –5 • 3n–1
jeka94

Answer:

<h2>C. -5; -135, -10,935</h2>

Step-by-step explanation:

\text{Substitute}\ n=1,\ n=4\ \text{and}\ n=8\ \text{to}\ A(n)=-5\cdot3^{n-1}:\\\\A(1)=-5\cdot3^{1-1}=-5\cdot3^0=-5\cdot1=-5\\\\A(4)=-5\cdot3^{4-1}=-5\cdot3^3=-5\cdot27=-135\\\\A(8)=-5\cdot3^{8-1}=-5\cdot3^7=-5\cdot2,187=-10,935

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Solve the equation ?
Morgarella [4.7K]
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3 years ago
Find the first five terms of the sequence given the following recursive formula:
forsale [732]

Answer:

The first five terms are;

-3,-7,11,-29,69

Step-by-step explanation:

The recursive definition of the sequence is

a_1=-3,, a_2=-7 and a_n=a_{n-2}-2a_{n-1}.

When n=3, we obtain;

a_3=a_{3-2}-2a_{3-1}.

\implies a_3=a_{1}-2a_{2}.

\implies a_3=-3-2(-7).

\implies a_3=11.

When n=4

\implies a_4=a_{2}-2a_{3}.

\implies a_4=-7-2(11)=-29.

When n=5

\implies a_5=a_{3}-2a_{4}.

\implies a_4=11-2(-29)=69.

Therefore the first five terms are;

-3,-7,11,-29,69

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3 years ago
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