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KiRa [710]
2 years ago
15

Can y’all please solve this

Mathematics
1 answer:
Wittaler [7]2 years ago
8 0
A). The answer is x = -4
Work:
3 - x = 11 + x
3 - x - 3 = 11 + x - 3
-x = x + 8
-x - x = x + 8 - x
-2x = 8
-2x/-2 = 8/-2
Answer: x = -4

B). The answer is x = 6
Work:
2x + 7 = 3x + 1
2x + 7 = 3x + 1 - 7
2x = 3x - 6
2x - 3x = 3x - 6 - 3x
-x = -6
-x/-1 = -6/-1
Answer x = 6

C). Answer: x = 4/3
Work:
8 - 2x = 4 + x
8 - 2x - 8 = 4 + x - 8
-2x = x - 4
-2x - x = x - 4 - x
-3x = -4
-3x/-3 = -4/-3
Answer: x = 4/3

D). Answer: x = 2
Work:
4x + 2 = 2x + 6
4x + 2 - 2 = 2x + 6 - 2
4x = 2x + 4
4x - 2x = 2x + 4 - 2x
2x = 4
2x/2 = 4/2
Answer: x = 2
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Can someone help me with this? PLs i'm so confused!
barxatty [35]
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}

13. U. sin\ A = \frac{a}{c} = \frac{hypotenuse}{opposite} = \frac{BC}{AB} = \frac{12}{15} = \frac{4}{5}

14. I. cos\A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{9}{15} = \frac{3}{5}

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

16. R. sin\ B = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{4}{\sqrt{65}} = \frac{4}{\sqrt{65}} * \frac{\sqrt{65}}{\sqrt{65}} = \frac{4\sqrt{65}}{65}

17. M. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{7}{4} = 1\frac{3}{4}

18. N. tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{4}{7}

19. L. sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{16}{34} = \frac{8}{17}

20. H. cos\ B = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \fac{AC}{AB} = \frac{30}{34} = \frac{15}{17}

21. tan\ B = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{16}{30} = \frac{8}{15}

22. O. sin\ A = \frac{a}{c} = \frac{opposite}{hypotenuse} = \frac{BC}{AB} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

23. O. cos\ A = \frac{b}{c} = \frac{adjacent}{hypotenuse} = \frac{AC}{AB} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

24. N. tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{1} = 1
7 0
3 years ago
Paul spends $89.10 on blackberries. If the blackberries cost $4.95 per pound, how many pounds of blackberries did he buy?
aksik [14]
18 pounds
U can divide 89.10 by 4.95 and u get 18
To get if it’s right multiple 18 by 4.95 and U get 89.10
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8.25% convert to a decimal then round to fourth decimal Place I
Amiraneli [1.4K]
As 100%=1.0

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3 years ago
1/10 of what number is 10 times as much as 8
Nataly_w [17]
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The table displays the distribution of blood types A, B, AB, and O to the presence of the Rh factor that is either present (Pos.
tino4ka555 [31]

Answer:

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