Hiroto's texting plan costs $20 per month, plus $0.05 per
2 answers:
The question is incomplete:
Hiroto's texting plan costs $20 per month, plus $0.05 per text message that is sent or received. Emilia's plan costs $10 per month and $0.25 per text. Using the graph below, which statement is true?
A. Hiroto’s plan costs more than Emilia’s plan when more than 50 texts are sent.
B. Both plans cost the same when 22 texts are sent.
C. Emilia’s plan costs more than Hiroto’s plan when more than 22 texts are sent.
D. Both plans cost the same when 50 texts are sent.
THE GRAPH IS ATTACHED.
Answer:
D. Both plans cost the same when 50 texts are sent.
Explanation:
The answer is that both plans cost the same when 50 texts are sent as both lines intersect in 50 texts with a price of $22.5 which means that they cost the same.
The other options are not right:
-Hiroto’s plan costs more than Emilia’s plan when more than 50 texts are sent. This is not true because in the graph you can see that after 50 texts Emilia's line is higher which indicates that her plan has a higher costs.
-Both plans cost the same when 22 texts are sent. This is not true because the lines don't intersect at the same point when 22 texts are sent.
-Emilia’s plan costs more than Hiroto’s plan when more than 22 texts are sent. This is false because when 22 texts are sent Emilia's plan costs around $15.5 and Hiroto's plan costs around $21 and Hiroto's plan is more expensive until they reach 50 texts that is the only point where they cost the same.
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Greetings!
"<span>What is the general process for solving an equation with one variable?"...
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<span>Add -4 to both sides to isolate the variable.
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Simplify.
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Divide both sides by 2 to isolate the variable on one side.
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"<span>What is the general process for solving an equation with one variable?"...
Typically when solving an equation with one variable, your objective is to isolate the variable on one side of the equation. This can be done adding/subtracting number to cancel them out on one side. You can also multiply/divide coefficients in order to isolate a variable.
Example:
</span>
<span>Add -4 to both sides to isolate the variable.
</span>
Simplify.
</span>
Divide both sides by 2 to isolate the variable on one side.
Hope this helps.
-Benjamin
<span>The rate of the express train is 15 mph faster than the freight train.
Let's grasp onto this piece of information.
The rate at which the freight train goes will be labeled "x" (keep in mind rate is just the speed)
If the freight train has a speed of "x" then the express train which is going 15mph faster than the freight train can be represented as "x + 15mph."
Our result is in HOURS, but our speed is in Miles per Hour (Miles/Hour). Meaning we need to multiply our speeds by some time (t).
Let's set up two equations:
Freight Train: 18 hours * x mph = distance
Express Train: 15 hours * (x mph + 15mph) = distance
They are both the same distance so we can set the two equations equal to each other like this:
18 hours * x mph = 15 hours * (x mph + 15mph)
That right side is a bit of a thorn in our side, so let's simplify it by multiplying all the terms in the parenthesis by 15 hours (keep in mind the hours in the units cancel out).
18 hours * x mph = 15x miles + 15 * 15 miles
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18x miles = 15x miles + 225 miles
Get all the "x" terms on one side by subtracting 15x on both sides.
18x - 15x = 225 miles
Simplify
3x = 225 miles
Divide by 3 on both sides to solve for x.
225/3
225 = 25*9
(25*9)/3 = 25*3 = 75mph
So "x" which we chose was the rate of the freight train is 75mph.
Adding 15mph to the freight train rate gives you the express trains rate which is 90mph.
Hope that helped.
Let's grasp onto this piece of information.
The rate at which the freight train goes will be labeled "x" (keep in mind rate is just the speed)
If the freight train has a speed of "x" then the express train which is going 15mph faster than the freight train can be represented as "x + 15mph."
Our result is in HOURS, but our speed is in Miles per Hour (Miles/Hour). Meaning we need to multiply our speeds by some time (t).
Let's set up two equations:
Freight Train: 18 hours * x mph = distance
Express Train: 15 hours * (x mph + 15mph) = distance
They are both the same distance so we can set the two equations equal to each other like this:
18 hours * x mph = 15 hours * (x mph + 15mph)
That right side is a bit of a thorn in our side, so let's simplify it by multiplying all the terms in the parenthesis by 15 hours (keep in mind the hours in the units cancel out).
18 hours * x mph = 15x miles + 15 * 15 miles
18 hours * x mph = 15x miles + 225 miles
Let's simplify the left side (</span>
18x miles = 15x miles + 225 miles
Get all the "x" terms on one side by subtracting 15x on both sides.
18x - 15x = 225 miles
Simplify
3x = 225 miles
Divide by 3 on both sides to solve for x.
225/3
225 = 25*9
(25*9)/3 = 25*3 = 75mph
So "x" which we chose was the rate of the freight train is 75mph.
Adding 15mph to the freight train rate gives you the express trains rate which is 90mph.
Hope that helped.