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3 years ago

Please answer please it’s pweety ez

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

8 0

Answer:

67 feet^2

Step-by-step explanation:

Surface Area = 2×(4×2.6 + 4×3.5 + 2.6×3.5) = 67 feet^2

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Natalija [7]

Calculate the difference, Reduce the fraction with 2, Multiply the numbers

(-10)^2+(7/2)^2-10

To raise a fraction to a power, raise the numerator and denominator to that power

100+49/4 -10

Calculate the sum

409/4

Or

102 1/4

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102.25

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3 years ago

Convert the following inches to feet and inches 39 inches 42.5 inches 61 3/4 inches

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39 inches = 3 feet 3 inches
42.5 inches = 3 feet 6 1/2 inches
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3 years ago

एउटा कार रु 18 लाख 5 सयमा किनेर 22 लाखमा बिक्री गर्दा कति प्रतिशत फाईदा हुन्छ

mars1129 [50]

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3 0

3 years ago

The rate of change of the downward velocity of a falling object is the acceleration of gravity (10 meters/sec 2) minus the accel

Alexus [3.1K]

Answer:

See below

Step-by-step explanation:

Write the initial value problem and the solution for the downward velocity for an object that is dropped (not thrown) from a great height.

if v(t) is the speed at time t after being dropped, v'(t) is the acceleration at time t, so the the initial value problem for the downward velocity is

v'(t) = 10 - 0.1v(t)

v(0) = 0 (since the object is dropped)

The equation v'(t)+0.1v(t)=10 is an ordinary first order differential equation with an integrating factor

$\bf e^{\int {0.1dt}}=e^{0.1t}$

so its general solution is

$\bf v(t)=Ce^{-0.1t}+100$

To find C, we use the initial value v(0)=0, so C=-100

and the solution of the initial value problem is

$\bf \boxed{v(t)=-100e^{-0.1t}+100}$

what is the terminal velocity?

The terminal velocity is

$\bf \lim_{t \to\infty}(-100e^{-0.1t}+100)=100\;mt/sec$

How long before the object reaches 90% of terminal velocity?

90% of terminal velocity = 90 m/sec

we look for a t such that

$\bf -100e^{-0.1t}+100=90\rightarrow -100e^{-0.1t}=-10\rightarrow e^{-0.1t}=0.1\\-0.1t=ln(0.1)\rightarrow t=\frac{ln(0.1)}{-0.1}=23.026\;sec$

How far has it fallen by that time?

The distance traveled after t seconds is given by

$\bf \int_{0}^{t}v(t)dt$

So, the distance traveled after 23.026 seconds is

$\bf \int_{0}^{23.026}(-100e^{-0.1t}+100)dt=-100\int_{0}^{23.026}e^{-0.1t}dt+100\int_{0}^{23.026}dt=\\-100(-e^{-0.1*23.026}/0.1+1/0.1)+100*23.026=1,402.6\;mt$

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3 years ago

What is the mean proportional between 5 and 6

Misha Larkins [42]

The answer is 5.5 I’m pretty sure

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3 years ago