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qwelly [4]
1 year ago
14

Below is the graph of y = |x| . .Translate it to make it the graph of of y=|x-2|+4

Mathematics
1 answer:
sineoko [7]1 year ago
7 0
Explanation

Given a function f(x) we translate the function:

• a units horizontally (a > 0 to the right, a < 0 to the left),

,

• b units vertically (b > 0 up, b < 0 down),

by the transformation:

f(x)\rightarrow g(x)=f(x-a)+b.

In this case, we have:

\begin{gathered} f(x)=|x|, \\ g(x)=|x-2|+4=f(x-2)+4. \end{gathered}

Comparing f(x) and g(x) with the general transformation above, we see that the graph of g(x) is the graph of f(x) translated:

• a = 2 units to the right,

,

• b = 4 units up.

Translating the graph of f(x), we get:

Answer

The translated graph is the graph in red:

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Answer:

x > 4

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Hope this helps.

6 0
3 years ago
Find the actual sum of 1 3/4+1 5/8 do the Turman how much David grew into years explain how you know your answer is reasonable
kotegsom [21]

Answer:

  3 3/8

Step-by-step explanation:

We assume your intended question is ...

  "Find the actual sum of 1 3/4 + 1 5/8 to determine how much David grew in two years. Explain ..." (Proper word choice and punctuation help communication.)

The sum of the two given mixed numbers can be found several ways. One way is to write the fractions using a common denominator:

  1 3/4 + 1 5/8 = 1 6/8 + 1 5/8 = (1 +1) +(6/8 +5/8) = 2 + 11/8 = 3 3/8

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3 0
3 years ago
1 - 5/6 &lt; 3x + 5/6 -2x
Romashka-Z-Leto [24]

Answer:

x>

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3

Step-by-step explanation:

6 0
3 years ago
Give an example of a function f : N → N that is surjective but not injective. You must explain why your example is surjective an
Helen [10]

Answer:

Consider f: N → N defined by f(0)=0 and f(n)=n-1 for all n>0.

Step-by-step explanation:

First we will prove that f is surjective. Let y∈N be any natural number. Define x as the number x=y+1. Then x∈N, and f(x)=x-1=(y+1)-1=y.  We conclude that f is surjective.

However, f is not injective. Take x1=0 and x2=1. Then x1≠x2 but f(x1)=0 and f(x2)=x2-1=1-1=0. We have shown that there are two natural numbers x1,x2 such that x1≠x2 but f(x1)=f(x2), that is, f is not injective.

Note:

If 0∉N in your definition of natural numbers, the same reasoning works with the function f: N → N defined by f(1)=1 and f(n)=n-1 for all n>1. The only difference is that you consider x1=1, x2=2 for the injectivity.

7 0
3 years ago
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