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2 years ago

What Is 1/8 of 3? someone please help

Mathematics

2 answers:

fredd [130]2 years ago

5 0

1/8 of 3 is 0.375

1/8 +3/1 = 3/8 = 0.375

maw [93]2 years ago

3 0

1/8 of 3 is 0.375 Hope it helps!

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In 4 days, the price of a share of stock rose 3/4 of a point. On a average, what was the change in stock each day? Answers: A) 3

jok3333 [9.3K]

Answer:

i think the answer is option 1 3

Step-by-step explanation:

i say this bc 3/4 times 4 is 3

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3 years ago

36√(3)-(432/7+4√(3))=?

Elden [556K]

40√3+432/7 is the answer

Hope this helps :D

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3 years ago

Find f(3) for f(x) = 1/2(4)^x <br> A. 8<br> B. 32<br> C. 6<br> D. 64

Elza [17]

1/2 (4)^3

1/2 x 64

32

7 0

3 years ago

Read 2 more answers

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

Please please please help me!!!!!!

soldier1979 [14.2K]

Answer:

Step-by-step explanation:

Use Pythagorean theorem.

c^2 = a^2+b^2

c^2 = 4^2+6^2

c^2 = 16+36

c^2 = 52

c = √52 = 7.21

7 0

3 years ago