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2 years ago

1 Point

Mathematics

1 answer:

Paha777 [63]2 years ago

5 0

Answer:

Step-by-step explanation:

$f(x)=5^{3x}$

Horizontal asymptote: y = 0

Exponential growth.

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What is the area of this figure

MAVERICK [17]

Answer:

Step-by-step explanation:

7-3=4

3x4=12/3=4

do that two times because there is two triangles

7x7=49

4+4+49=57

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3 years ago

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Figure LMNO is a parallelogram.<br> What is the value of x?

galina1969 [7]

Values is 6 for lmso is a parellelle gram

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3 years ago

15% of $9 is $ this is a question on my math worksheet please respond quick and show the steps on how to do it please and thank

Vladimir79 [104]

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2 years ago

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Last question of my homework pls <3

dimaraw [331]

Answer:

J(2, 5 )

Step-by-step explanation:

Using the section formula

$x_{J}$ = $\frac{1(-2)+2(4)}{2+1}$ = $\frac{-2+8}{3}$ = $\frac{6}{3}$ = 2

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Thus J = (2, 5 )

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

3 years ago