How many solutions can be found for the equation 5x + 3(x − 1) = 10x − 2x − 3?

1 answer:

3 0

Infinitely many solutions. Since your slope and Y axis values are the same on both sides no matter what you fill in for x your answer will be true.

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Which are steps that could be used to solve 0 = 9(x2 + 6x) – 18 by completing the square? Check all that apply.

jeyben [28]

18 + 81 = 9(x²<span> + 6x + 9)
</span><span>11 = (x + 3)</span>²

When we are completing the square, we are going to move the value of c across the equals. We will do that by adding, and end up with

18=9(x²+6x)

We take the value of b (the coefficient of x), divide it by 2 and square it:
(6/2)²=3²=9

This is the value that completes the square. However, since the entire square is multiplied by 9, this value must be multiplied by 9 before we can add it across the equals:
18+9(9) = 9(x²+6x+9)
18+81=9(x²+6x+9)
99=9(x²+6x+9)

Dividing both sides by 9, we have:
11=x²+6x+9

11=(x+3)²

8 0

3 years ago

Read 2 more answers

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$