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2 years ago

(1over5^2(3+2+5)^2=i need helllllllllllllllllp pleaaaaaaassssseee

Mathematics

2 answers:

Nikitich [7]2 years ago

8 0

Answer:

2500

Step-by-step explanation:

5^2(3+2+5)^2

5^2(10)^2

25(100)

=2500

Step2247 [10]2 years ago

6 0

Answer:

Step-by-step explanation:

becuz smart

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Elenna [48]

Step-by-step explanation:

part A

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part B

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(f(x2) - f(x1)) / (x2 - x1)

so,

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Answer:

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Step-by-step explanation:

example : 3 ± 5 = 8 or -2

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2 years ago

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Calculate the rectangular prism with a length of 5.6cm, a with of 2.1 cm, and a height of 6.6cm.

Inessa05 [86]

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2 years ago

HELPPP PLEASE THANKS

neonofarm [45]

Answer:

ur dumb kid

Step-by-step explanation

learn it your self

7 0

2 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

2 years ago

(1over5^2(3+2+5)^2=i need helllllllllllllllllp pleaaaaaaassssseee​

(1over5^2(3+2+5)^2=i need helllllllllllllllllp pleaaaaaaassssseee