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Effectus [21]
3 years ago
7

Ram’s father is 38 years old. His age is three more than five times the age

Mathematics
2 answers:
den301095 [7]3 years ago
8 0

Answer:

7

Step-by-step explanation:

7 times 5 =35

35+3=38

correct answer is 7 years old

Naddik [55]3 years ago
4 0

Answer:

Ram is 7

Step-by-step explanation:

38-3

=35 divided by 5

=7

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Which ordered pair is a solution to the system?
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A.

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3 years ago
A square pyramid is shown: 0.4 feet , 0.4 feet , 5 feet What is the surface area of the pyramid?
Vlad [161]
A square pyramid is called like so because its base is a square.

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\bf \stackrel{\textit{squarish base}}{(0.4\cdot 0.4)}~~+~~\stackrel{\textit{4 triangles}}{4\left[\cfrac{1}{2}(0.4)(5)  \right]}\implies 0.16~~+~~4\implies 4.16
5 0
3 years ago
Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.
finlep [7]

Answer:

a) v = \frac{[L]}{[T]} = LT^{-1}

b) a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

c) \int v dt = s(t) = [L]=L

d) \int a dt = v(t) = [L][T]^{-1}=LT^{-1}

e) \frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

v = \frac{ds}{dt}

a= \frac{dv}{dt}

Part a

If we do the dimensional analysis for v we got:

v = \frac{[L]}{[T]} = LT^{-1}

Part b

For the acceleration we can use the result obtained from part a and we got:

a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

Part c

From definition if we do the integral of the velocity respect to t we got the position:

\int v dt = s(t)

And the dimensional analysis for the position is:

\int v dt = s(t) = [L]=L

Part d

The integral for the acceleration respect to the time is the velocity:

\int a dt = v(t)

And the dimensional analysis for the position is:

\int a dt = v(t) = [L][T]^{-1}=LT^{-1}

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

7 0
3 years ago
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