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Yuki888 [10]
3 years ago
8

Solve with substitution5-y=3x y=2x​

Mathematics
1 answer:
vodomira [7]3 years ago
4 0

<h2> <u>y= − 3 x + 5 </u></h2>

Isolate the variable by dividing each side by factors that don't contain the variable.

Step-by-step explanation:

Thats the answer of your 5-y=3×

hopes this helps you Correct me if I'm wrong:=)

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What is the effect on the graph of the function f(x) = x^2 when f(x) is changed to f(x − 6)
olya-2409 [2.1K]

Answer:

The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing f(x) = x^2 and \:f\left(x\right)=\left(x-6\right)^2 is represented as being shifted 6 units to the right as compare to the function f(x) = x^2.

Step-by-step explanation:

When we Add or subtract a positive constant, let say c, to input x, it would be a horizontal shift.  

For example:

Type of change               Effect on y = f(x)

y = f(x - c)                      horizontal shift: c units to right

So

Considering the function

f(x) = x^2

The graph is shown below. The first figure is representing f(x) = x^2.

Now, considering the function

\:f\left(x\right)=\left(x-6\right)^2

According to the rule, as we have discussed above, as a positive constant 6 is added to the input, so there is a horizontal shift, 6 units to the right.

The graph of \:f\left(x-6\right)=\left(x-6\right)^2 is shown below in second figure. It is clear that the graph of  \:f\left(x-6\right)=\left(x-6\right)^2  is shifted 6 units to the right as compare to the function f(x) = x^2.

The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing f(x) = x^2 and \:f\left(x\right)=\left(x-6\right)^2 is represented as being shifted 6 units to the right as compare to the function f(x) = x^2.

5 0
3 years ago
Need help ASAP BRAINLIEST IF CORRECT
kiruha [24]

Answer:

X=6

Step-by-step explanation:

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V125BC [204]

Answer:

D.\ \angle FOB\ and\ \angle EOC

Step-by-step explanation:

\angle FOC=180\textdegree\ \ \ \ \ \ \ \ \ \ as\ FOC\ is\ a\ straight\ line\\\\\angle FOC=\angle COE+\angle EOF\\\\\angle COE+\angle EOF=180\textdegree\\\\Hence\ \angle COE\ is\ supplementary\ to\ \angle EOF

\angle EOB=180\textdegree\ \ \ \ \ \ \ \ \ \ as\ EOB\ is\ a\ straight\ line\\\\\angle EOB=\angle FOB+\angle EOF\\\\\angle FOB+\angle EOF=180\textdegree\\\\Hence\ \angle FOB\ is\ supplementary\ to\ \angle EOF

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