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3 years ago

A submarine wa 525 feet below sea level. it ascends up 140 feet what is the position

Mathematics

2 answers:

Ivenika [448]3 years ago

4 0

Answer: It will be 385 feet below the sea after the ascension

Step-by-step explanation:

Gennadij [26K]3 years ago

3 0

Answer:

the sub is 585 feet below the surface after it ascends 140 feet from it's original position. hope that helps

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Help ASAP Given the function f(x) = 3x + 1, what is the value of f(-3)

vlada-n [284]

Answer:

-8

Step-by-step explanation:

f(x) = 3x + 1

f(-3) = 3(-3) + 1

F(-3)=-9+1

f(-3)= -8

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3 years ago

How do you solve this problem 35/8y+1491-2y

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Answer:

I suggest using the website called Cymath https://www.cymath.com/

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Step-by-step explanation:

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3 years ago

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Một ủy ban an toàn kiểm tra ngẫu nhiên 900 công nhân xây dựng trong thời gian làm việc, thấy có 48 công nhân không mang mũ bảo h

Usimov [2.4K]

Answer:

The percentage of workers not wearing the helmets is 5.3 %.

Step-by-step explanation:

A safety committee randomly examined 900 construction workers during their work, and found that 48 workers were not wearing helmets. Estimate the percentage of workers who do not wear protective masks during their working time with 98% confidence

total workers = 900

Not wearing helmet = 48

Percentage which are not wearing the helmets

= $\frac{48}{900}\times 100 = 5.3 %$ %

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3 years ago

The volume of a volleyball is about 288 cubic inches. what is the radius of the volleyball, to the nearest tenth of an inch? use

Len [333]

Answer:

The radius of the volleyball is 8.3 inches

Step-by-step explanation:

Given

$r = \sqrt{\frac{3v}{4\pi}}$

$v = 288$

Required

Determine the value of r

To do this, we simply substitute 288 for v and 3.14 for π in the given equation.

This gives

$r = \sqrt{\frac{3v}{4\pi}}$

$r = \sqrt{\frac{3 * 288}{4 * 3.14}}$

$r = \sqrt{\frac{864}{12.56}}$

$r = \sqrt{68.7898089172}$

$r = 8.29396219651$

$r = 8.3\ inch$ (Approximated)

Hence;

The radius of the volleyball is 8.3 inches

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3 years ago

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A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739

OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
Then for the x direction it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, for the y direction, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$

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3 years ago