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VMariaS [17]
3 years ago
9

what is the effect on the graph of the linear parent function, f (x) = x, when f ( x ) is replaced by f (x +5)

Mathematics
1 answer:
satela [25.4K]3 years ago
3 0

Answer:

When f(x) is replaced by f(x+5), it will shift the parent function '5 units' to the left.

Step-by-step explanation:

  • We know that when we add a number 'a' to the input of the function, it would move the parent function 'a' units to the left.

In other words, the rule is:

  • f(x + a) will shift the parent function 'a units' to the left.

Given the function

f(x)=x

Thus, when f(x) is replaced by f(x+5), it will shift the parent function '5 units' to the left.

  • The effect on the graph of the linear parent function is shown in the attached diagram.

In the graph, the red line is representing the parent function f(x) and the blue line is representing the effect on the graph i.e. f(x+5).

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3.1 - 2.6 = 0.5 pounds
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Can anyone pls help me in this pls
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(x + y)² and z² + 4(1/2 xy) both represent the area of the outer square and are equal.

Step-by-step explanation:

  • Step 1: The below reasons explain why the expression is a true equation.

1. The expression (x + y)² finds the area of the outer square by squaring its side length.

[In the figure, side length of the outer square = x + y and area of a square = (side length)²]

2. The expression z² + 4(1/2 xy) finds the area of the outer square by adding the area of the inner square and the four triangles.

[In the figure the length of the inner square is z, the triangles have a base x and height y and area of a triangle = 1/2 base × height]

⇒ So the left hand side and right hand side of the expression is equal.

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3 years ago
I just don’t get it I keep getting a negative
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Answer:

30

Step-by-step explanation:

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4 years ago
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If a sprinkler waters 1 over 15 of a lawn in 1 over 3 hour, how much time will it take to water the entire lawn?
Feliz [49]
Now, a whole, will be 15/15 or 1 whole.

now, the sprinkler does 1/15 of the lawn in 1/3 of an hour, how long will it be to do the whole 15/15?

\bf \begin{array}{ccll}
lawn&hours\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
\frac{1}{15}&\frac{1}{3}\\\\
\frac{15}{15}&h
\end{array}\implies \cfrac{\quad \frac{1}{15}\quad }{\frac{15}{15}}=\cfrac{\quad \frac{1}{3}\quad }{h}\implies \cfrac{1}{15}\cdot \cfrac{15}{15}=\cfrac{\quad \frac{1}{3}\quad }{\frac{h}{1}}
\\\\\\
\cfrac{1}{15}=\cfrac{1}{3}\cdot \cfrac{1}{h}\implies \cfrac{1}{15}=\cfrac{1}{3h}\implies 3h=15\implies h=\cfrac{15}{3}\implies h=5
6 0
3 years ago
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Larry and Paul start out running at a rate of 5 mph. Paul speeds up his pace after 5 miles to 10 mph but Larry continues the sam
klasskru [66]
<h2>Hello!</h2>

The answer is:

They will be 10 miles apart after 3 hours.

<h2>Why?</h2>

To calculate how long after they start will they be 10 miles apart, we need to assume that after 1 one hour, they were at the same distance (5 miles), then, calculate the time when they are 10 miles apart, knowing that Paul increased its speed two times, running first at 5mph and then, at 10 mph.

The time that will pass to be 10 miles apart can be calculated using the following equation:

TotalTime=TimeToReach5miles+TimeToBe10milesApart

Calculating the time to reach 5 miles for both Larry and Paul, at a speed of 5 mph, we have:

x=xo+v*t\\\\5miles=0+5mph*t\\\\t=\frac{5miles}{5mph}=1hour

We have that to reach a distance of 5 miles, they needed 1 hour. We need to remember that at this time, they were at the same distance.

If we want to know how many time will it take for them to be 10 miles apart with Paul increasing its speed to 10mph, we need to assume that after that time, the distance reached by Paul will be the distance reached by Larry plus 10 miles.

So, for the second moment (Paul increasing his speed) we have:

For Larry:

x_{L}=5miles+5mph*t

Therefore, the distance of Paul will be equal to the distance of Larry plus 10 miles.

For Paul:

x{L}+10miles=xo+10mph*t\\\\5miles+5mph*t+10miles=5miles+10mph*t\\\\5miles+10miles-5miles=10mph*t-5mph*t\\\\10miles=5mph*t\\\\t=\frac{10miles}{5mph}=2hours

Then, there will take 2 hours to Paul to be 10 miles apart from Larry after both were at 5 miles and Paul increased his speed to 10 mph.

Hence, calculating the total time, we have:

TotalTime=TimeToReach5miles+TimeToBe10milesApart

TotalTime=1hour+2hours=3hours

Have a nice day!

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