Answer:
The y-intercept of the equation is 100 and represents the initial studio-use fee.
Step-by-step explanation:
In this equation, our t variable (time) is the equivalent of the x-variable on a graph. This is because it is the variable that we 'change' to see its impact on y. We see how the amount of hours affects the price. So our P variable (price) is the equivalent of y on a graph. The y-intercept is where the line crosses the y-axis on a graph. At this point, x=0.
Since P is our y, and t is our x, to find the y-intercept, we simply need to make t = 0.
P = 50(0) + 100
P = 100
Therefore the y-intercept is 100.
In this context, t represents time, so even though the studio has been used for 0 hours, the price is still 100. This is because the 100 represents the initial studio-use fee, and using it for certain amounts of time adds onto the initial fee of $100. The hourly fee is represented by 50t so it costs $50 more for each hour of use.
Hope this helped!
1. We assume, that the number 118 is 100% - because it's the output value of the task.
<span>2. We assume, that x is the value we are looking for. </span>
<span>3. If 118 is 100%, so we can write it down as 118=100%. </span>
<span>4. We know, that x is 47% of the output value, so we can write it down as x=47%. </span>
5. Now we have two simple equations:
1) 118=100%
2) x=47%
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
118/x=100%/47%
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.
7. Solution for what is 47% of 118
118/x=100/47
<span>(118/x)*x=(100/47)*x - </span>we multiply both sides of the equation by x
<span>118=2.12765957447*x - </span>we divide both sides of the equation by (2.12765957447) to get x
<span>118/2.12765957447=x </span>
<span>55.46=x </span>
x=55.46
<span>now we have: </span>
<span>47% of 118=55.46</span>
It is 22 my guy because i think it is 22
Wow. this is too difficult for me
Answer:
The answer is B! All I did was take a screenshot of your question and put on my page.
