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3 years ago

Write an equation for each translation of .

Mathematics

1 answer:

alexdok [17]3 years ago

7 0

Answer:

Step-by-step explanation:

y = f(x) + n - moves the graph n units up

y = f(x) - n - moves the graph n units down

y = f(x + n) - moves the graph n units to the left

y = f(x - n) - moves the graph n units to the right

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

5.5 units to the right

y = f(x - 5.5) = |x - 5.5|

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Someone help me please.

Maru [420]

Answer:

Step-by-step explan

1 x 4/2 + 1

4/2 + 1

4 divide by 2 is = 2

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so the value of 1 x b/2 + 1 is = to 3

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4 years ago

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How many times is the image reflected in a 6-sided snowflake?

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3 years ago

One way to show that two triangles are similar is to show that

avanturin [10]

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show that the three sets of corresponding sides are in proportion. If the three sets of corresponding sides of two triangles are in proportion, the triangles are similar.

Step-by-step explanation:

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3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago