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3 years ago

Please help me solve this

Mathematics

2 answers:

kicyunya [14]3 years ago

6 0

Answer:

The answer is 700.4 cm.

Step-by-step explanation:

Hope it helped (ik im a bit late but i tried)

Diano4ka-milaya [45]3 years ago

5 0

Answer:

l=700.4 cm

Step-by-step explanation:

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madam [21]

Answer: the answer is D. the spread is from 1 to 6.

Step-by-step explanation:

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3 years ago

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In the figure below, one side of the right triangle is a diameter of the semicircle.

olga55 [171]

Answer:

Option (A)

Step-by-step explanation:

One side of the given triangle is a diameter of the semicircle given.

Measure of the diameter = 10 units

Total area of the semicircle = $\frac{1}{2}\pi (r^{2})$

= $\frac{1}{2}\pi (5)^2$

= 39.27 square units

Area of the right triangle = $\frac{1}{2}(\text{Base})(\text{Height})$

= $\frac{1}{2}(6)(8)$

= 24 square units

Area of the shaded region = Area of the semicircle - Area of the right triangle

= 39.27 - 24

= 15.27 square units

≈ 15 square units

Therefore, option (A) will be the answer.

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3 years ago

3 60° х x = - y=-- How to solve

DochEvi [55]

Answer5

Step-by-step explanation:

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2 years ago

When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v squared w

Sati [7]

Answer:

Step-by-step explanation:

The model fo the shell is given by the following equation of equilibrium:

$\Sigma F = -b\cdot v^{2} - m\cdot g = m\cdot \frac{dv}{dt}$

This first-order differential equation has separable variables, which are cleared herein:

$\int\limits^t_{0\,s} \, dt = -\frac{m}{b} \int\limits^{0\,\frac{m}{s} }_{600\,\frac{m}{s} } {\frac{1}{ v^{2}+\frac{m}{b}\cdot g } } \, dv$

The solution of this integral is:

$t = -\frac{m}{2b}\cdot \left[\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } }\right) - \tan^{-1} \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }\right)\right]$

$\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } } \right)=-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)$

$\frac{v}{\sqrt{\frac{m\cdot g}{b} } }=\tan \left[-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\right]$

$v = \sqrt{\frac{m\cdot g}{b} } \left [\frac{\tan \left(-\frac{2\cdot b \cdot t}{m} \right)+ \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)}{1 - \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\cdot \tan \left(-\frac{2\cdot b \cdot t}{m} \right) }\right]$

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3 years ago

What is the product of 0.4 and 2.5

Maksim231197 [3]

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