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3 years ago

Help Please!!!!! Evaluate the expression when n = 4 2n-7

Mathematics

2 answers:

Montano1993 [528]3 years ago

7 0

2•4= 8 8-7= 1
The answer is 1

crimeas [40]3 years ago

6 0

Answer:

the answer is 1.

Step-by-step explanation:

4 is equal to n so it would look like 24-7 but you have to multiply 2 and 4 which is 8. Then, you do 8-7 which is 1. hope this helped.

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Dave works two jobs. He works at the grocery store making $12 per hour and works at the car wash making $15 per hour. If he work

vladimir2022 [97]

Answer:

11 hours

Step-by-step explanation:

let he works at first store x hours and at car wash y hours. Then

x+y=25

multiply by 4

4x+4y=100 ...(1)

12x+15y=333

divide by 3

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3 years ago

35 multiplied by the quantity r less than 45

Reil [10]

Hello There!

Call the missing value 'n'
35n < 45.

Hope This Helps You!
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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:

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

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