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3 years ago

Find the 61st term of the arithmetic sequence

Mathematics

1 answer:

krek1111 [17]3 years ago

5 0

Answer:

Step-by-step explanation:

hello :

a(n) -a(n-1) = -16 (common differnce for arithmetic sequence)

-44-(-28) = (-28)-(-12) = -16

the ni-eme term is : a(n) = a(1) +(n-1)d d = -16 and a(1) = -12

so : a(n) = -12 +(n-1)(-16)

a(n) = -16n+4

continue for 61st term n = ?????

when : n= ? a(?) = -16(?)+3 =.....

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3 years ago

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3 years ago

Erik and Caleb were trying to solve the equation: 0=(3x+2)(x-4) Erik said, "The right-hand side is factored, so I'll use the zer

Mumz [18]

Answer:

C) Both

Step-by-step explanation:

The given equation is:

$0=(3x+2)(x-4)$

To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.

Using this property:

$(3x+2) = 0 \Rightarrow \bold{x = -\frac{2}{3}}\\(x-4) = 0 \Rightarrow \bold{x = 4}$

So, Erik's solution strategy would work.

Now, let us discuss about Caleb's solution strategy:

Multiply $(3x+2)(x-4)$ i.e. $3x^2-12x+2x-8$ = $3x^2-10x-8$

So, the equation becomes:

$0=3x^2-10x-8$

Comparing this equation to standard quadratic equation:

$ax^2+bx+c=0$

a = 3, b = -10, c = -8

So, this can be solved using the quadratic formula.

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

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The answer is same from both the approaches.

So, the correct answer is:

C) Both

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