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aivan3 [116]
3 years ago
12

5x^2-41x+8=0 please helpp

Mathematics
2 answers:
jeka943 years ago
7 0

Answer:

Use the quadratic formula

=

−

±

2

−

4

√

2

x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}

x=2a−b±b2−4ac​​

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

5

2

−

4

1

+

8

=

0

5x^{2}-41x+8=0

5x2−41x+8=0

=

5

a={\color{#c92786}{5}}

a=5

=

−

4

1

b={\color{#e8710a}{-41}}

b=−41

=

8

c={\color{#129eaf}{8}}

c=8

=

−

(

−

4

1

)

±

(

−

4

1

)

2

−

4

⋅

5

⋅

8

√

2

⋅

5

2

Simplify

3

Separate the equations

4

Solve

Solution

=

8

=

1

5

prisoha [69]3 years ago
3 0
X=8
X= 1/5 is solution
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A sequence is defined by the formula f(n+1)=f(n)-3. If f(4)=22, what is f(1)?
Vlad [161]
<h3>Answer: Choice C) 31</h3>

==============================================

Explanation:

The recursive rule

f(n+1)=f(n)-3

can be rearranged to

f(n) = f(n+1)+3

after adding 3 to both sides

----------------

Now let's say we plug in n = 3

f(n) = f(n+1)+3

f(3) = f(3+1)+3

f(3) = f(4)+3

f(3) = 22+3

f(3) = 25

Repeat for n = 2

f(n) = f(n+1)+3

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f(2) = f(3)+3

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3 years ago
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