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garri49 [273]
3 years ago
7

5x-y=-15 x+y=-3 USE SUBSTITUTION METHOD

Mathematics
2 answers:
Kobotan [32]3 years ago
7 0

Answer:

5x - y = -15

x + y = -3   ===>  y = -3 -x


5x - (-3 - x) = -15

5x + 3 - x = -15

5x - x = -15 -3

4x = -18

x = -18/4

x = -4.5


x + y = -3

-4.5 + y = -3

y = -3 + 4.5

y = 1.5




Rzqust [24]3 years ago
5 0

What I got was x= -3 and y=0

5x-y=-15

x+y=-3

y=-3-x

5x-(-3-x)=-15 (solve for X)

x=-3

y=-3-(-3)

y=0

(-3,0)

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1. E. sine\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{5}{13}

2. L. cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{12}{13}

3. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{5}{12}

4. Y. sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{5}{13}

5. W. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{12}{13}

6. tan\ B = \frac{b}{a} = \frac{adjacent}{opposite} = \frac{AC}{BC} = \frac{12}{5} = 2\frac{2}{5}

7. sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{2}

8. W. cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{\sqrt{3}}{2}

9. I. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}

10. sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{2}

11. E. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{\sqrt{3}}{1} = \sqrt{3}

12. I. tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}

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24. N. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{1} = 1
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