All categories

2 years ago

PLZ answer correctly! Brainliest!

Mathematics

1 answer:

polet [3.4K]2 years ago

5 0

Answer:

$A=\left(-1\dfrac{5}{5}\right)=(-2)\\\\B=\left(-\dfrac{4}{5}\right)\\\\C=\left(\dfrac{3}{5}\right)$

Step-by-step explanation:

Look at the picture.

$\dfrac{2}{10}=\dfrac{1}{5}$

You might be interested in

NEED ANSWER ASAPPPPPP

Oksana_A [137]

The volume is 1017.88
If you search up “how to solve a cone” you can get the same answer

6 0

2 years ago

Read 2 more answers

Mike is mixing paint for his walls. He mixes 1/6 gallon blue paint and 5/8 gallon green paint in a large container. What fractio

Reptile [31]

1/6 + 5/8 =
8/48 + 30/48 = 38/48 reduces to 19/24 <==== total gallons of paint

8 0

3 years ago

A rectangular swimming pool is 25 ft long and 20 1/2 ft wide, and it must be filled to a depth of 4 1/2 ft. One cubic foot of wa

bazaltina [42]

ANSWER

307½ gallons of water

EXPLANATION

The volume of the rectangular swimming pool is

$V = l \times w \times h$

where l=25ft and w=20½ ft.

when it is filled to a depth of 4½ ft ,then the height is , h=4½ ft.

We substitute these values into the formula to get:

$V = 25\times 20.5\times 4.5$

$V = 2306.25 {ft}^{3}$

If one cubic foot of water is 7½ gallons, then 2306.25 cubic feet will be

$\frac{2306.25}{7.5} = 307.5gallons$

We need 307.5 gallons of water to fill the pool.

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

If you built a model of the Empire State Building that is proportional to the real building. The height of the Empire State Buil

tangare [24]

Let’s see now, there are 600 inches in 50 feet.
12in = 1ft, 12in * 50 = 600.
Which means the model you’ll be making will be a 1:600 ratio model. If we find out how many inches are in 1,250 feet, and divide that number by 600, we will get our answer.
1,250 * 12 = 15,000.
So 1,250 feet equate to 15,000 inches. Now we must divide 15,000 by 600.
15,000 ÷ 600 = 25.
So your model will be 25 inches (or 2’1” [2 feet & 1 inch] tall).

7 0

3 years ago