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VARVARA [1.3K]
4 years ago
12

Write the equations after translating the graph of y = |x|: one unit up,

Mathematics
1 answer:
Tatiana [17]4 years ago
5 0

Answer:

g(x) = |x| + 1

Step-by-step explanation:

Given

y = |x|

Required

Translate 1 unit up

Start by replacing y with f(x)

f(x) = |x|

To translate an the graph of an absolute function upward, you make use of the formula;

g(x) = f(x) + k

Where k is the number of units

In this case; k = 1

Hence;

g(x) = f(x) + k

Substitute k = 1

g(x) = f(x) + 1

Substitute  f(x) = |x|

g(x) = |x| + 1

<em>Hence, the resulting equation is </em>g(x) = |x| + 1<em></em>

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Mumz [18]

Answer:

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are\ in\ AP

Step-by-step explanation:

Given that (b-c)^2, (c-a)^2 , (a-b)^2 are in AP

To prove: \dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are in AP

From given as we know if p , q, r are in AP then 2q= p+r.

2(c-a)^2= (b-c)^2+(a-b)^2\\\\\Rightarrow 2(c^2+a^2-2ac)=b^2+c^2-2bc+a^2+b^2-2ab\\\\\Rightarrow 2c^2+2a^2-4ac= 2b^2+c^2+a^2 -2bc-2ab\\\\\Rightarrow a^2+c^2-2b^2-4ac= -2bc-2ab\\\\\Rightarrow a^2-2b^2+c^2= 4ac-2bc-2ab

Now

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}2\dfrac{1}{(c-a)} =\dfrac{1}{(b-c)}+\dfrac{1}{(a-b)}\\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-b+b-c}{(b-c)(a-b)} \\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-c}{(b-c)(a-b)} \\\\\Rightarrow2(b-c)(a-b) = (c-a)(a-c) \\\\\Rightarrow 2(ab-b^2-ac+bc)= -(a-c)^2\\\\\Rightarrow 2ab- 2b^2-2ac+2bc = -a^2-c^2+2ac\\\\\Rightarrow a^2-2b^2+c^2=4ac-2ab-2bc

Which is the result of AP

.

Hence proved

6 0
3 years ago
Can someone help me with this problem
scoray [572]

Answer:

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Step-by-step explanation:

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4 years ago
Consider the following binomial experiment: A study in a certain community showed that 5% of the people suffer from insomnia. If
svp [43]

Answer:

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Step-by-step explanation:

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3 years ago
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