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Natasha2012 [34]
2 years ago
15

Help and explain !!!!( Solve using the substitute method )

Mathematics
1 answer:
alex41 [277]2 years ago
5 0

Answer:

x=13

y= -9

(13, -9)

Step-by-step explanation:

If we are solving using the substitution method, we can take the first equation and set it to y so y=4-x.

Then, we can take that equation and plug it into the bottom one so

3x+4(4-x)=3

Simplify:

3x+16-4x=3

-x=-13

x=13

We can then plug 13 into any of the two given equations (I am just going to plug it into the top one)

So, 13+y=4 which y= -9

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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}

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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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15. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{12}{9} = \frac{4}{3} = 1\frac{1}{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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