1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
scZoUnD [109]
3 years ago
14

Which of the following is the solution to the equation x^2-36x=0

Mathematics
2 answers:
Marizza181 [45]3 years ago
7 0
x^2-36x=0 \\
x(x-36)=0 \\
x=0 \ \lor \ x-36=0 \\
x=0 \ \lor \ x=36 \\
\boxed{x=0 \hbox{ or } x=36}
Inessa05 [86]3 years ago
3 0
x^2-36x=0\\
x(x-36)=0\\
x=0 \vee x=36
You might be interested in
4) y =<br> Justify:<br> 63°<br> у
kenny6666 [7]

Answer:

y = 63° because they are alternate angles.

3 0
2 years ago
I need help finding y
kolezko [41]

to solve for y, we must multiply by t on both side to get ride of the t in the denominator. By doing this, we will get:

{t}^{3}

on the right side.

Subtracting 3, and we have successfully isolated y.

It would be impossible to get a quantitative value for y if we don't know the value of t.

4 0
3 years ago
Read 2 more answers
Tell me how you solved thanks
Allushta [10]
A. Similar triangles have equal angles even though they're not the same size that was similar means So then

X^2-5x=4x+36
X^2-5x-4x-36=0
X^2-9x-36=0
(X-12)(x(+3)
X=12. X=3

Angle C and angle F are equal
So 4(12)-5=43
4 0
3 years ago
Solve for X and y<br> 60°<br> 14 3
lisov135 [29]

Answer:

X = 28

Y = 14

Step-by-step explanation:

+++++++++++++++++++++++++++++++++++++

4 0
2 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
Other questions:
  • Is the product of 10 over 3 bigger than 54
    5·1 answer
  • What is the starting point of creating a argument on triangle congruence?
    9·1 answer
  • Answer please need help with this
    6·1 answer
  • X+ =4+2y=5 solve this​
    14·2 answers
  • Plot triangle ABC with coordinates A(-4,5),B(-1,5),C(-1,1)​
    13·1 answer
  • Please answer this quickly! Thank you!
    10·1 answer
  • Which expression can you simplify by combining like terms?
    6·1 answer
  • A craft show has an admission fee of $1.50 for children and $4.00 for adults. On Saturday 155 people came to the carnival and $5
    11·1 answer
  • Does anyone know the answer to question 3
    5·1 answer
  • There are 20 pieces of fruit in a basket and 12 of them are oranges what percentage of the pieces of fruit in the basket or oran
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!