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3 years ago

What type of parent function does the equation f (x) = 1/x

Mathematics

1 answer:

mr_godi [17]3 years ago

6 0

Answer:

idk.. that's a good question

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Solve the below: 4x+3y=7 3x-2y=9

Ahat [919]

4x+3y=7
3x-2y=9

12x+9y=21
12x-8y=36
^^Multiply top equation by 3 and multiply bottom equation by 4

12x+9y=36
- 12x-8y=36

17y=-15
^^Then subtract those two new equations

17y/17= -15/17

Y= -0.88
^^Next divide both sides by 17

4x+3* -0.88=7

4x + (-2.64) =7
-(-2.64) -(-2.64)

4x = 9.64

4x/4=9.64/4
X=2.41

^^then choose on equation and put your new y in the y spot and solve.

X=2.41

Y=-0.88

Your point/coordinate is (2.41,-.88)

7 0

3 years ago

How can you write 0.35 in a fraction

Vladimir [108]

.35 as a fraction is 35/100=7/20 Hope this helps :D

5 0

3 years ago

Read 2 more answers

NEED THIS ASAPWrite the standard form of the equation of each line given the slope and y-intercept

Mama L [17]

Note:

Your first question is missing the y-intercept, so I am solving the 2nd question. You would still get your concept clear because the procedure to solve each of the questions is the same.

Question 2

Answer:

The equation in the standard form is:

2/3x + y = -4

Please also check the attached graph.

Step-by-step explanation:

We know that the equation in the standard form is

Ax + By = C

where x and y are variables and A, B and C are constants

Given

Slope m = -2/3

y-intercept = -4

To determine

Write the equation in the standard form

We know that the slope-intercept form of the line equation

$y = mx+b$

where

m is the slope

b is the y-intercept

In our case:

Slope m = -2/3

y-intercept = -4

substituting m = -2/3 and y-intercept b = -4 in the slope-intercept of the line equation

y = mx+b

y = -2/3x + (-4)

y = -2/3x - 4

Writing the equation in the standard form

2/3x + y = -4

Therefore, the equation in the standard form is:

2/3x + y = -4

Please also check the attached graph.

4 0

3 years ago

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

Solve the inequality. 2 < 6 - 3r and graph the solution

Vladimir79 [104]

the answer is 3 because you subtract 6 - 3 = 3r

6 0

3 years ago