g(x) = 3√(x-5) -1
The process of altering a graph to produce a different version of the preceding graph is known as graph transformation. The graphs can be moved about the x-y plane or translated. They may also be stretched, or they may undergo a mix of these changes.
Horizontal stretching: It means the graph is elongated or shrink in x direction.
Vertical stretching : It means the graph is elongated or shrink in y direction
Vertical translation : It means moving the base of the graph in y direction
Horizontal translation : It means moving the base of the graph in x direction
According to rules of transformation f(x)+c shift c units up and f(x)-c shift c units down.
Therefore, in order to move the graph down 1 units, we need to subtract given function by 1 , we get
g(x) = 3√x -1
According to rules of transformation f(x+c) shift c units left and f(x-c ) shift c units right.
Therefore, in order to move the graph left by 5 units, we need to add given function by 5 , we get
g(x) = 3√(x-5) -1
To learn more about graphical transformation, refer to brainly.com/question/4025726
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Answer:
4 different ways
Step-by-step explanation:
Total number of children = 4
Distribution of the 4 children :
Number of boys = 3 ; Number of girls = 1
Boy = B ; Girl = G
Possible combinations :
BBBG ; GBBB ; BBGB ; BGBB
From the pascal triangle number of e; number of outcomes = 2
Having exactly 3 boys and 1 girl
Hence, of any of the 4 four total children, 3 must be boys and 1 girl ;
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
![\sqrt[4]{ x^{3} } = x^{3/4}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%20x%5E%7B3%7D%20%7D%20%3D%20x%5E%7B3%2F4%7D%20)
![\sqrt[10]{ x^{5} * x^{4}* x^{2} } = \sqrt[10]{ x^{11} }= x^{11/10}](https://tex.z-dn.net/?f=%20%5Csqrt%5B10%5D%7B%20x%5E%7B5%7D%20%2A%20x%5E%7B4%7D%2A%20x%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%5B10%5D%7B%20x%5E%7B11%7D%20%7D%3D%20x%5E%7B11%2F10%7D%20%20)
