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Kitty [74]
3 years ago
6

(please help)How many real solutions exist for this system of equations? y = x2 + 4 y = 4x

Mathematics
1 answer:
astra-53 [7]3 years ago
6 0

Answer:

B

Step-by-step explanation:

y=x^(2)+4

y=4x

lets solve by substitution

4x=x^2+4

0=x^2-4x+4

0=(x-2)(x-2)

x=2 or x=2

y=4x

y=4*2

y=8

same for the other answer

(2,8)

this system has only one solution: (2,8)

Thus, the correct answer is B (one solution)

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Answer:

<u>This isn't really that hard to solve.</u>

Step-by-step explanation:

For the first one (H): <em>3 units to the left of 0</em>.

This has to mean that our integer will be negative, and thus, -3 should be written. To the right of 0 would mean it's positive. So, you would find the spot for -3 on the number line and write the letter "H."

The last one (H): <em>not positive or negative</em>.

This means that the integer would have to be 0, as it satisfies the condition that the integer is neither positive (+) nor negative (-). So, you would find the spot marked 0 on the number line and write the letter "H."

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Answer:

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Step-by-step explanation:

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