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3 years ago

What is 5(3-12)(42-32)

Mathematics

2 answers:

laiz [17]3 years ago

8 0

Answer:

-450

Step-by-step explanation:

5(3-12)(42-32)

5(-9)(10)

-45(10)

Sati [7]3 years ago

6 0

Answer:

-450

Step-by-step explanation:

5(3-12)(42-32)

5(-9)(10)

-45(10)

= -450

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Which is another way of correctly expressing<br> (x - 3)(x - 3)?

worty [1.4K]

(x-3)^2, as x-3 is repeated twice

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3 years ago

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How many times smaller is the surface area of a cube if the side length is multiplied 1/2<br>

Tju [1.3M]

Answer:

1/4, or 4 times smaller

Step-by-step explanation:

Since the surface area of a cube is just the combination of the area of all of the square sides, and the formula for those is s^2, if the side length was decreased by 1/2 then the surface area would decrease by (1/2)^2=1/4. Hope this helps!

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3 years ago

Find the slope of the line passing through the points (-2,5) and (3/2, 2).

sveticcg [70]

Answer:

The answer is

Step-by-step explanation:

The slope of the line passing through two points can be found by using the formula

where

(x1 , y1) and (x2 , y2) are the points

From the question

The points are (-2,5) and (3/2, 2)

The slope is

<h3> $m = \frac{2 - 5}{ \frac{3}{2} + 2} \\ = - \frac{3}{ \frac{7}{2} } \\ = - 3 \div \frac{7}{2} \\ = - 3 \times \frac{2}{7}$ </h3>

We have the final answer as

Hope this helps you

8 0

3 years ago

dexters weighs 50 kg. His body contains 30.8 kg of oxygen and 4. 9 kg if hydrogen. The rest is made up of other elements. How ma

andrey2020 [161]

50-(30.8+4.9)=14.3 kg.
You start with 50 then you need to subtract 30.8 and 4.9 from it. So you add them together and get 35.7. Then you subract 25.9 from 50 and get 14.3 kg.

4 0

4 years ago

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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