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3 years ago

A solid figure has 5 faces, 8 edges, and 4 vertices. What figure is it?

2 answers:

8 0

I'm not really sure but the square pyramid has 5 faces 8 edges 5 vertices and the triangular pyramid has 4 faces 8 edges 4 vertices I'm not sure what the correct answer is but hopefully this helps you

Ganezh [65]3 years ago

5 0

It should possibly be the square pyramid that has 5 faces.

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You have a job and you make $11.45 per hour. If your total check for the week was $171.75, how many hours did you work this week

zaharov [31]

Answer: you have worked 15 hours

Step-by-step explanation: i multiplied $11.45 and 15 which got me to $171.75 your welcome ;)

6 0

2 years ago

Order from least to greatest<br> 14,20,15,10,7,12,14,15,10

defon

Answer:

7 10 10 12 14 14 15 15 20

3 0

3 years ago

Read 2 more answers

I just need these 4, they should be super easy!! I really need help

valkas [14]

Answer:

1. v = 72

2. b = 8

3. 0 or -2

4. C = -24

Step-by-step explanation:

Hope this helps.. Can I get brainiest

7 0

2 years ago

Question 1 (1 point)

Artemon [7]

Answer:

the answer is simply the one in the bottom right, B

3 0

2 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