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3 years ago

What is the relationship between the Two 3s in 453,389?

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

5 0

If you multiply the 300 by 10 it becomes 3000

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Tomio, age 28, takes out $50,000 of straight-life insurance. His annual premium is $418.20. Using the tables found in the textbo

olya-2409 [2.1K]

<span>Tomio, age 28, takes out $50,000 of straight-life insurance. His annual premium is $418.20. Using the tables found in the textbook, determine the cash value of his policy at the end of 20 years.

A. $26,500
B. $30,000
C. $13,250
D. $26,000

</span>50,000/1,000 = 50
265*50 = $13,250

So the cash value of this policy is $13,250
<span>
The correct answer is:
</span><span>C. $13,250</span>

6 0

3 years ago

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What number is added to 0.034 to obtain 9.079?

grigory [225]

9.045 is the number added to 0.034 to obtain 9.079

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3 years ago

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What is the total surface area of the rectangular pyramid below

Novosadov [1.4K]

Answer:

Step-by-step explanation:

We have

the rectangle 12ft × 6ft

two triangles with the base b = 12ft and the height h = 8ft

two triangles with the base b = 6ft and the height h = 9.5ft.

The formula of an area of a rectangle l × w:

$A=lw$

Substitute:

$A_1=(12)(6)=72\ dt^2$

The formula of an area of a triangle:

$A=\dfrac{bh}{2}$

Substitute:

$A_2=\dfrac{(12)(8)}{2}=48\ ft^2$

$A_3=\dfrac{(6)(9.5)}{2}=28.5\ ft^2$

The Surface Area:

$S.A.=A_1+2A_2+2A_3$

Substitute:

$S.A.=72+2(48)+2(28.5)=225\ ft^2$

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3 years ago

Read 2 more answers

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. $tan\ A = \frac{a}{b} = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{1}{1} = 1$

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3 years ago

Explain how the number of pieces in a whole relates to the size of each piece

astra-53 [7]

As the number of pieces in a whole gets larger, the size of the pieces gets smaller. This is represented by the denominator in a fraction. As the denominator increases, the number of pieces the whole is divided into also increases so the size of the pieces is smaller.
1/2 > 1/3 > 1/4 > 1/5 etc...

Fractions get smaller as the denominator gets bigger.

4 0

3 years ago