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Mashutka [201]
3 years ago
13

If f(x)=x^2 is vertically compressed by a factor of 8 yo g(x) , what is the equation of g(x)

Mathematics
1 answer:
Wewaii [24]3 years ago
5 0

Answer:

g(x)=\frac{1}{8} x^{2}

Step-by-step explanation:

Expansions and compressions are transformations that change the length or width of the graph of a function. The general form of the graph of a function expands or compresses vertically or horizontally. Expansions and compressions are considered non-rigid transformations.

<u><em>Vertical compression:</em></u>

<u><em></em></u>

<u><em></em></u>y=kf(x)<u><em></em></u>

<u><em></em></u>

  • If k>0 the graph of f (x) is vertically stretched by a factor of k.

  • If k<0 the vertical stretch is followed by a reflection across the x-axis.

  • If 0<k<1 the graph is f (x) vertically compressed by a factor of 1/k.

So, if f(x)=x^{2}  is vertically compressed by a factor of 8. Hence, using the previous information, we can conclude that the equation of g(x) is:

g(x)=\frac{1}{8} x^{2}

I attached you the graphs.

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8 0
3 years ago
Please Solve<br><br> 6(x + 4) = 30
lianna [129]

Answer: x=1

Step-by-step explanation: 6x+24=30

6x=6

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6 0
3 years ago
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A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?
alexandr1967 [171]

Answer:

The probability that the last card dealt is an ace is \frac{1}{13}.

Step-by-step explanation:

Given : A deck of ordinary cards is shuffled and 13 cards are dealt.

To find : What is the probability that the last card dealt is an ace?

Solution :

There are total 52 cards.

The total arrangement of cards is 52!.

There is 4 ace cards in total.

Arrangement for containing ace as the 13th card is 4\times 51!.

The probability that the last card dealt is an ace is

P=\frac{4\times 51!}{52!}

P=\frac{4\times 51!}{52\times 51!}

P=\frac{4}{52}

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Therefore, the probability that the last card dealt is an ace is \frac{1}{13}.

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Answer:

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Step-by-step explanation:

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7 0
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find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the large
katrin2010 [14]

A generic odd number can be written as

2k+1,\quad k \in \mathbb{Z}

Since there is an odd number every two numbers, three consecutive odd numbers will be

2k+1,\quad 2k+3,\quad 2k+5

Now let's make up the equations: the sum of the first two is

(2k+1)+(2k+3)

And 27 less than 3 times the largest is

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These two must be the same, so we have

(2k+1)+(2k+3)=3(2k+5)-27 \iff 4k+4 = 6k+30-27 \iff 4k+4=6k+3

Subtracting 4k and 3 from both sides gives

1=2k \iff k=\dfrac{1}{2}

Which means that the problem has no solution.

To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even

On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).

So, the left hand side is even, and the right hand side is odd. They can't be the same number.

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