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2 years ago

A road is inclined at an angle of 3°. After driving 5330 feet along this road, find the drivers increase in altitude

Mathematics

1 answer:

Wewaii [24]2 years ago

8 0

Hope this helps just some notes

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Please help with math and explain if u do i’ll give brainly!!

Nikolay [14]

Answer:

3/2

Step-by-step explanation:

MNOP to ABCD: The answer has to be greater than one because ABCD increases in size from MNOP.

To get the scale factor you have to compare the sides.

Line AD = 6 in.

Line MP = 4 in.

6/4 = 3/2

3 0

2 years ago

A certain three-cylinder combination lock has 60 numbers on it. To open it, you turn to a number on the first cylinder, then

statuscvo [17]

Answer:

(a) The number of different lock combinations is 216,000.

(b) The probability of getting the correct combination in the first try is $\frac{1}{216000}$ .

Step-by-step explanation:

The lock has three-cylinder combinations with 60 numbers on each cylinder.

The procedure of the opening lock is to turn to a number on the first cylinder, then to a second number on the second cylinder, and then to a third number on the third cylinder.

The numbers can be repeated.

(a)

There are three cylinder combinations on the lock, each with 60 numbers.

It is provided that repetitions are allowed.

Then each cylinder can take any of the 60 numbers.

So there are 60 options for the first cylinder.

There are 60 options for the second cylinder.

And there are 60 options for the third cylinder.

So the total number of possible combinations is:

Total number of combinations = 60 × 60 × 60 = 216000.

Thus, the number of different lock combinations is 216,000.

(b)

The events of getting a correct combinations of the three-cylinder combination lock implies that all the three cylinder are set at the correct numbers.

Each cylinder has 60 numbers.

This implies that there are 60 possible ways to get a correct number for the first cylinder.

The probability of getting the correct number for the first cylinder is:

P (Number on 1st cylinder is correct) = $\frac{1}{60}$ .

Similarly for the second cylinder the probability of getting the correct number is:

P (Number on 2nd cylinder is correct) = $\frac{1}{60}$ .

And similarly for the third cylinder the probability of getting the correct number is:

P (Number on 3rd cylinder is correct) = $\frac{1}{60}$ .

So the probability of getting the correct combination in the first try is:

P (Correct combination in the 1st try) = $\frac{1}{60}\times \frac{1}{60}\times \frac{1}{60}=\frac{1}{216000}$ .

Thus, the probability of getting the correct combination in the first try is $\frac{1}{216000}$ .

7 0

2 years ago

Make vthe subjject of the formulaE=mvsquare/ 2 find the value of v when m =2 and E=64

Maksim231197 [3]

E=mv²

divide both sides by m

E/m=v²

square root both sides

v=√E/m

m=2,E=64

v=√64/2

v=9/√2

6 0

2 years ago

What is the volume of this sphere?

Natasha_Volkova [10]

Answer:

Option c. 11,488.21 cm3

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The volume of the sphere is equal to

$V=\frac{4}{3}\pi r^{3}$

we have

$r=28/2=14\ cm$ ---> the radius is half the diameter

$\pi =3.14$

substitute

$V=\frac{4}{3}(3.14)(14)^{3}$

$V=11,488.21\ cm^3$

3 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

A road is inclined at an angle of 3°. After driving 5330 feet along this road, find the drivers increase in altitude​

A road is inclined at an angle of 3°. After driving 5330 feet along this road, find the drivers increase in altitude