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Y_Kistochka [10]
3 years ago
12

Consider the graph of the linear function h(x) = –x + 5. Which could you change to move the graph down 3 units? the value of b t

o –3 the value of m to –3 the value of b to 2 the value of m to 2
Mathematics
2 answers:
storchak [24]3 years ago
8 0

Answer:

B.

Step-by-step explanation:

Consider the graph of the linear function h(x) = –x + 5. Which could you change to move the graph down 3 units?

the value of b to –3  

the value of m to –3  

the value of b to 2

the value of m to 2

Naetoosmart
2 years ago
its C
Artist 52 [7]3 years ago
3 0
The value of b to 2.
By doing that you are applying a vertical translation to the function
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Answer:

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Step-by-step explanation:

Required

Select graphs that are dilated by a scale factor greater than 1

For graph A:

Graph A is smaller than the original graph. This indicates dilation with a scale factor less than 1

For graph B:

Graph B is bigger than the original graph and is dilated over (0,0). This indicates dilation with a scale factor greater than 1

For graph C:

Graph C is bigger than the original graph; however, it is not dilated over (0,0). This indicates dilation with a scale factor greater than 1

For graph D:

Graph D is bigger than the original graph; however, it is not only dilated but also flipped over (i.e. rotated).

<em>Hence, b and c is true</em>

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Step-by-step explanation:


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Please help logarithms!
nlexa [21]

Given:

\log_34\approx 1.262

\log_37\approx 1.771

To find:

The value of \log_3\left(\dfrac{4}{49}\right).

Solution:

We have,

\log_34\approx 1.262

\log_37\approx 1.771

Using properties of log, we get

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349      \left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2      

\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37          [\log x^n=n\log x]

Substitute \log_34\approx 1.262 and \log_37\approx 1.771.

\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)

\log_3\left(\dfrac{4}{49}\right)=1.262-3.542

\log_3\left(\dfrac{4}{49}\right)=-2.28

Therefore, the value of \log_3\left(\dfrac{4}{49}\right) is -2.28.

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