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Y_Kistochka [10]
2 years ago
14

Solve 15 ≥ -3x or 2/5x ≥ -2.

Mathematics
1 answer:
Lera25 [3.4K]2 years ago
8 0

Answer:

15 \geqslant  - 3x  \\  \frac{15}{3}  \geqslant  - x \\ 5 \geqslant  - x \\  - 5 \leqslant x

\frac{2}{5}x \geqslant  - 2 \\ x  \geqslant  - 2 \times  \frac{5}{2}   \\ x \geqslant  - 5

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Geometry question for homework.
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So remember that the distance formula is \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} .

1.

\sqrt{(4-4)^2+(-2-(-5))^2}\\ \sqrt{0^2+3^2}\\ \sqrt{0+9}\\ \sqrt{9}\\ 3

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3 years ago
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And there you have it

Hope it helps!

8 0
3 years ago
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