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3 years ago

Which is a valid proportion? 3 over 4 = 21 over 28

Mathematics

1 answer:

dangina [55]3 years ago

5 0

Answer: I think that it is the first one

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Y = 3x^2 - 4x<br> 27 + y = 14x

Hatshy [7]

Step-by-step explanation:

move the 27 to the other side of the equation like this

$y = 14x - 27$

use this equation and substitute it into the y value of the first equation

4 0

3 years ago

A family member has some five dollar bills and one dollar bills in their wallet. Altogether she has 18 bills and a total of $62.

Pepsi [2]

X - the number of five dollar bills
y - the number of one dollar bills

She has 18 bills in total.
$x+y=18 \\ y=18-x$

She has $62.
$5x+y=62 \\
y=62-5x$

Set both expressions equal to each other:
$18-x=62-5x \\
-x+5x=62-18 \\
4x=44 \\
x=11 \\ \\ y=18-x=18-11=7$

She has 11 five dollar bills and 7 one dollar bills.

7 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)

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

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

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

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