Y=f(x)=(4)^x find f(x) when x=1/2

SELECT ALL The Points that are on the Graph of the line 3x + 2y = 28 (:

Oksana_A [137]

<h2>Answer:</h2>

<h2>Step-by-step explanation:</h2>

points are on the graph of the line:-

$\tt{3x+2y=28 }$ ⠀

we get,

$\rightsquigarrow$ $\tt{ 2y=28-3x }$ ⠀

$\rightsquigarrow$ $\tt{y=\dfrac{28-3x}{2} }$ ⠀

In,(0,14)

If we put the value ,x=0

y= $\tt{\dfrac{ 28-3×0}{2}=\dfrac{28}{2}=14 }$

So,we get one point,(0,14)✔

In,(2,11)

If we put ,x=2

y= $\tt{\dfrac{28-3×2}{2}=\dfrac{28-6}{2}=\dfrac{22}{2}=11 }$

so,we get another point (2,11)✔

in other given ponts{(1,12),(4,9),(8,4)},It is impossible. ⠀ ⠀

5 0

3 years ago

Read 2 more answers

Dont really understand how to do this

bija089 [108]

Answers:

$t_{10} = -22 \ \text{ and } S_{10} = -85$

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

Explanation:

$t_1 = \text{first term} = 5\\t_2 = \text{first term}-3 = t_1 - 3 = 5-3 = 2$

Note we subtract 3 off the previous term (t1) to get the next term (t2). Each new successive term is found this way

$t_3 = t_2 - 3 = 2-3 = -1\\t_4 = t_3 - 3 = -1-3 = -4$

and so on. This process may take a while to reach $t_{10}$

There's a shortcut. The nth term of any arithmetic sequence is

$t_n = t_1+d(n-1)$

We plug in $t_1 = 5 \text{ and } d = -3$ and simplify

$t_n = t_1+d(n-1)\\t_n = 5+(-3)(n-1)\\t_n = 5-3n+3\\t_n = -3n+8$

Then we can plug in various positive whole numbers for n to find the corresponding $t_n$ value. For example, plug in n = 2

$t_n = -3n+8\\t_2 = -3*2+8\\t_2 = -6+8\\t_2 = 2$

which matches with the second term we found earlier. And,

$tn = -3n+8\\t_{10} = -3*10+8\\t_{10} = -30+8\\t_{10} = \boldsymbol{-22} \ \textbf{ is the tenth term}$

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

The notation $S_{10}$ refers to the sum of the first ten terms $t_1, t_2, \ldots, t_9, t_{10}$

We could use either the long way or the shortcut above to find all $t_1$ through $t_{10}$ . Then add those values up. Or we can take this shortcut below.

$Sn = \text{sum of the first n terms of an arithmetic sequence}\\S_n = (n/2)*(t_1+t_n)\\S_{10} = (10/2)*(t_1+t_{10})\\S_{10} = (10/2)*(5-22)\\S_{10} = 5*(-17)\\\boldsymbol{S_{10} = -85}$