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agasfer [191]
3 years ago
15

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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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
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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}

x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times3 \times (-8)}}{2\times 3}\\x=\dfrac{-(-10)\pm\sqrt{196}}{6}\\x=\dfrac{10\pm14}{6} \\\Rightarrow x= 4, -\dfrac{2}{3}

The answer is same from both the approaches.

So, the correct answer is:

C) Both

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