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2 years ago

The four 4's chalenge, 1-20

Mathematics

1 answer:

tiny-mole [99]2 years ago

5 0

4, 8, 12, 16, 20 this is the fours up to 20

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Been a while since I’ve needed to not use a calculator. Why is -2X10^-18 (3/16) = 4.09X10^-19?

Anestetic [448]

Answer:

see the explanation

Step-by-step explanation:

we have

$2.18*10^{-18}(\frac{3}{16})$

we have that

$\frac{3}{16}=0.1875$

Convert to scientific notation

$0.1875=1.875*10^{-1}$

substitute

$2.18*10^{-18}(1.875*10^{-1}})$

Remember that

To multiply two numbers in scientific notation, multiply their coefficients and add their exponents

$2.18*10^{-18}1.875*10^{-1}}=(2.18*1.875)*10^{(-18-1)}=4.0875*10^{-19}$

Round the number 4.0875 two decimal places

$4.0875*10^{-19}=4.09*10^{-19}$

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3 years ago

You are situated 300 feet from the base of Tower Glitz Plaza watching an external elevator descend down the side of the building

nordsb [41]

Answer:

18.66 ft/s

Step-by-step explanation:

The distance between you and the elevator is given by:

$h=\sqrt{x^2+y^2}$

The rate of change for the distance between you and the elevator is given by:

$\frac{dh}{dt}=\frac{dh}{dy}*\frac{dy}{dt}$

$-16=\frac{dh}{dy}*\frac{dy}{dt}$

$\frac{dh}{dy}=\frac{d}{dy} (\sqrt{x^2+y^2})\\$

Applying the chain rule:

$u=x^2+y^2\\\frac{dh}{dy}=\frac{d\sqrt u}{du} *\frac{du}{dy}\\\frac{dh}{dy}=\frac{1}{2\sqrt u} *2y\\\frac{dh}{dy}=\frac{y}{\sqrt {(x^2+y^2)}}$

Therefore, at x=300 and y = 500, dy/dt is:

$-16=\frac{y}{\sqrt {(x^2+y^2)}}*\frac{dy}{dt}\\-16=\frac{500}{\sqrt {(300^2+500^2)}}*\frac{dy}{dt}\\\frac{dy}{dt}=-18.66\ ft/s$

The elevator is descending at 18.66 ft/s.

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3 years ago

Complete the proof for proving that the diagonals of an isosceles trapezoid are congruent.

Alina [70]

..............................[a+b]2+c2

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3 years ago

I need help ASAP, with numbers 3-6 above. Please help.

MatroZZZ [7]

Answer:

3. 2 6 30 6. 8:5

3 9 45

Step-by-step explanation:

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3 years ago

What is the range of the function f(x) = |x| + 5?

Ierofanga [76]

Answer:

$[0, \infty)$

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1 year ago

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