All categories

3 years ago

Round to the nearest cent $1.6845

Mathematics

2 answers:

vova2212 [387]3 years ago

5 0

Answer:

$1.68

Step-by-step explanation:

Remember that cents are in the hundredths place. So think of it as rounding to the nearest hundredth place. 1.6845 ---> 1.68

Ede4ka [16]3 years ago

3 0

Answer:

$1.68

Step-by-step explanation:

old rhyme i learnt when i was little but it helps with rounding:

5 or more let it soar, 4 or less let it rest

You might be interested in

Plz help I’m getting timed

bixtya [17]

Answer:

17in.

Step-by-step explanation:

7 0

3 years ago

Elisa is writing an algebraic expression for the phrase 5 less than a number

Alekssandra [29.7K]

The algebraic expression can be written as
x-5.

3 0

3 years ago

3x^2 (x^2 - 4x - 2) + (5x^3 + 7x^2+2x+11) plz show work

matrenka [14]

Let's simplify step-by-step.

3x2(x2−4x−4)+5x3+7x2+2x+11

Distribute:

=(3x2)(x2)+(3x2)(−4x)+(3x2)(−4)+5x3+7x2+2x+11

=3x4+−12x3+−12x2+5x3+7x2+2x+11

Combine Like Terms:

=3x4+−12x3+−12x2+5x3+7x2+2x+11

=(3x4)+(−12x3+5x3)+(−12x2+7x2)+(2x)+(11)

=3x4+−7x3+−5x2+2x+11

Answer:

=3x4−7x3−5x2+2x+11

5 0

3 years ago

A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume t

sladkih [1.3K]

Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

$\frac{dy}{dt}\ \alpha\ y(1-y)$

$\frac{dy}{dt}=ky(1-y)$

where k is the constant of proportionality, dy/dt = rate at which the rumor spreads

$\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}} =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\$

$y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}$

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

$y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days$

5 0

3 years ago

PLEASE SOMEONE DO THE DAY7 : TIME FOR ME. GIVING BRAINLIEST FOR WHOEVER GETS IT RIGHT!

Blababa [14]

1. 1 hour and 30 minutes 2. 11:35 : your welcome !

8 0

3 years ago