Achurch rings its bells every 15 minutes, the school rings its bells every 20 minutes and the day care center rings its bells every 25 minutes. if they all ring their bells at noon on the same day, at what time will they next all ring their bells together? (answer in the form ab: cd without am or pm such as 08: 00)

5 : 00 PM

Step-by-step explanation:

Given,

Church ring bells every 15 minutes,

School rings its bells every 20 minutes,

And, day care center rings its bells every 25 minutes,

Thus, the number of minutes after which they will ring together = LCM (15, 20, 25)

∵ 15 = 3 × 5,

20 = 2 × 2 × 5,

25 = 5 × 5,

Thus, LCM (15, 20, 25) = 300

Now, 1 minute = 1/60 hours ⇒ 300 minutes = 5 hours,

Hence, after 5 hours they will ring together,

If they rang at 12:00 PM,

Then they will ring after 5 hours that is 5:00 PM.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.