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MAXImum [283]
2 years ago
10

John is 36 years old now. 1/4 of his age is equal to 1/3 of sam's age. find their total age in 5 years time

Mathematics
1 answer:
sveticcg [70]2 years ago
7 0

Answer:

John = 41 years

Sam = 32 years

Step-by-step explanation:

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I had to put two screenshot because it would screenshot the whole question but plz help
Tcecarenko [31]

Answer:

i think its D dont fullly take my word tho

Step-by-step explanation:

3 0
2 years ago
The function f(x) = (x - 4)(x-2) is shown.
ella [17]

Answer:

the range for f(x) = (x-4) (x-2) is [-1,∞)

Step-by-step explanation:

all real numbers greater than or equal to -1

4 0
2 years ago
Bill and Felicia each bake cookies for a party. Bill bakes 3 times as many cookies as Felicia. Felicia bakes 24 fewer cookies th
denis-greek [22]

Answer:

  • b = 3f
  • f = b -24

Step-by-step explanation:

The word problem describes two relationships between the numbers of cookies baked. When b and f represent the numbers of cookies baked by Bill and Felicia, respectively, those relationships are ...

  1. Bill bakes 3 times as many cookies as Felicia. (b = 3f)
  2. Felicia bakes 24 fewer cookies than Bill. (f = b-24)
4 0
3 years ago
Will someone help me work through this problem, please
Ainat [17]

Step-by-step explanation:

Radius (r) = 2 ft

V = 4/3 × 22/7 × (2)³

= 4/3 × 22/7 × 8

= 33.52 ft³

6 0
2 years ago
Inverse laplace of [(1/s^2)-(48/s^5)]
Katen [24]
**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform F(s) and G(s), then \mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}

Given that \dfrac{1}{s^2} = \dfrac{1!}{s^2} and -\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}

From Table of Laplace Transform, you have \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}} and hence \mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}.

Hope this helps...
7 0
3 years ago
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