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Natasha_Volkova [10]
3 years ago
7

please help me i don’t know how to do this. if you do know, please explain how you did it bc i have so many more of these to do,

ty!:)

Mathematics
1 answer:
Alex73 [517]3 years ago
6 0

Answer:

Blue: y = x² + 3

Purple: y = (x + 4)²

Step-by-step explanation:

The parabolas are "transformed", meaning they have been shifted on the coordinate grid. All parabolas will have the equation of "y = x²", any additional transformation will add onto that.

Blue: Looking at the red parabola, its lowest point is only three units down from the blue's lowest points. Meaning the blue parabola went only three units up. So you only have to do the "+ 3" in the equation.

Purple: Like in blue, we're gonna look at the purple's lowest point and compare it to red's lowest point. The purple parabola seems to have shifted only four units to the left. Though, instead of adding onto the equation like in the blue parabola, any changes on the x-axis will be put into parenthesis. IN ADDITION, the sign will be the opposite of what it is on the coordinate grid. For example, the purple parabola is on the -4 of the x-axis. Instead of putting "(x - 4)", we put "(x + 4)".

Try the next two problems on your own. They will be similarly structured to the two answers above. If you need more help, lemme know. :)

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5 0
3 years ago
Attorney A charges a fixed fee on $250 for an initial meeting and $150 per hour for all hours worked after that. Write an equati
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let h be the number of hours then

A → C = 150h + 250 ( where C is charge )

B → C = 175h + 150

for 26 hours

A → C = (150 × 26 ) + 250 = 3900 + 250 = $4150

B → C = (175 × 26 ) + 150 = 4550 + 150 = $4700

Attorney A is cheaper for 26 hours, thus better deal

Equate the 2 equations to find hours they charge the same

175h + 150 = 150h + 250 ( subtract 150h from both sides )

25h + 150 = 250 ( subtract 150 from both sides )

25h = 100 ( divide both sides by 25 )

h = 4 ← number of hours when charges for both are equal

Thus Attorney A becomes a better deal at 5 hours


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