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3 years ago

12

Each statement describes a transformation of the graph of f(x) = x. Which statement correctly describes the graph of g(x) if g(x

) = f(x) - 3?
A.
It is the graph of f(x) translated 3 units down.
B.
It is the graph of f(x) translated 3 units to the left.
C.
It is the graph of f(x) translated 3 units up.
D.
It is the graph of f(x) where the slope is decreased by 3.

2 answers:

torisob [31]3 years ago

5 0

Answer:

A

Step-by-step explanation:

Scorpion4ik [409]3 years ago

4 0

Answer:

A. It is the graph of f(x) translated 3 units down.

Step-by-step explanation:

I got it right in a test.

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The accompanying stem-and-leaf plot represents Kiko's test scores this year. What is the median score for this set of data?

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Write down all the values and cross out until you get to the middle, since there are 2 values, find the mean, which would be 80

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Evaluate −8x+5−2x−4+5x when x=2

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−8x+5−2x−4+5x

-8(2)+5-2(2)-4+5(2)

(-16)+5-4-4+10

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3 years ago

Lori want to distribute 35 peaches equally into baskets. she will use more than 1 but fewer than 10 baskets how many basket Lori

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Unless this is a trick question, I think 7 baskets. If you have 7 baskets then you can put 5 peaches in each basket. Hope this is right.

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A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Alex_Xolod [135]

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$

The number of seat when n is even $S_n=\frac{n^2+2n}{4}$

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[a+l]$

a = first term of the series.

d= common difference.

n= number of term

l= last term

$n^{th}$ term of a A.P series is

$T_n=a+(n-1)d$

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=1 and d=2

$n=1+(t-1)2$

⇒(t-1)2=n-1

$\Rightarrow t-1=\frac{n-1}{2}$

$\Rightarrow t = \frac{n-1}{2}+1$

$\Rightarrow t = \frac{n-1+2}{2}$

$\Rightarrow t = \frac{n+1}{2}$

Last term l= n,, the number of term $=\frac{ n+1}2$ , First term = 1

Total number of seat

$S_n=\frac{\frac{n+1}{2}}{2}[1+n}]$

$=\frac{{n+1}}{4}[1+n}]$

$=\frac{(1+n)^2}{4}$

$=\frac{n^2+2n+1}{4}$

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=2 and d=2

$n=2+(t-1)2$

⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$ , First term = 2

Total number of seat

$S_n=\frac{\frac{n}{2}}{2}[2+n}]$

$=\frac{{n}}{4}[2+n}]$

$=\frac{n(2+n)}{4}$

$=\frac{n^2+2n}{4}$

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Step-by-step explanation:

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