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3 years ago

The 6th term of an arithmetic progression is 35 and the 13th term is 77. find the 1st term

Mathematics

1 answer:

GenaCL600 [577]3 years ago

8 0

Answer:

Step-by-step explanation:

The 6th term of an arithmetic progression is 35 and the 13th term is 77. find the 1st term

We know we have an arithmetic sequence.

a_n = (n - 1)*d + a_1

given:

a_6 = 35 = (6-1)*d + a_1

a_13 = (13 -1)*d + a_1 = 77

find a_1

we have 35 = 5d + a_1

and 77 = 12d + a_1

2 equations in 2 unknowns.

We can solve this.

77 = 12d + a_1

35 = 5d + a_1

minus the 2 equations from each other.

77 - 35 = 12d - 5d + a_1 - a_1

42 = 7d

d = 6

Find a_1

35 = 5*6 + a_1

35 = 30 + a_1

a_1 = 35 -30 = 5

a_1 = 5

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Which equation is the inverse of y = 9x2 - 4?

m_a_m_a [10]

Answer:

The inverse will be:

$y' = \frac{\sqrt{x+4}}{3}$

Step-by-step explanation:

In order to find the inverse of the equation, we do a variable change, since we are finding the inverse, :

$f(x) = 9x^{2} - 4$

$y = 9x^{2} - 4$

$x = 9y' ^{2} - 4$

Now solve for y'.

First add 4 in both sides of the equation and change to the left y'.

$x + 4= 9y'^{2} - 4+4$

$9y'^{2}$ = x + 4

Second divide by 9

$9y'^{2}$ /9 = (x + 4)/9

$y'^{2}$ = (x + 4)/9

Now you will have to clear y, with the square root.

$[tex]y'^{\frac{2}{2}} = \sqrt{x + 4} / \sqrt{9}$ [/tex] =

Simplifying terms

$y' = \frac{\sqrt{x+4}}{3}$

$f^{-1}(x) = \frac{\sqrt{x+4}}{3}$

You can check the answer by doing the evaluation of the following equation:

(f o $f^{-1}$ ) (x)

substitute the equation for y' or inverse function $f^{-1}$

f( $\frac{\sqrt{x+4} }{3}$ )

Now substitue the value into f(x)

You will have

$= 9(\frac{\sqrt{x+4} }{3}} )^{2} - 4\\\\Solving\\\\9(\frac{{x+4} }{9}} ) - 4$

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