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3 years ago

15

Rafael counted a total of 40 white cars and yellow cars there are nine times as many white cars as yellow cars how many white ca

rs did Rafael count

1 answer:

natta225 [31]3 years ago

5 0

The answer probably 16 I don't really know

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If f(x) = -7x + 5x2 + 10, what does f(-2) equal?

Iteru [2.4K]

Answer:

44

Step-by-step explanation:

f(x) = -7x + 5x^2 + 10

f(-2) = -7(-2) + 5(-2)^2 + 10

= 14 + 20 + 10

= 44

6 0

3 years ago

Mandy charges all of her customers by weight so she gets $5 per kilogram of corn cobs or $1.25 for an individual corn cob. If Ma

guapka [62]

Answer:

$5

Step-by-step explanation:

Because everyone want to be rich!

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2 years ago

The rationalizing factor of 1/√13 is

SashulF [63]

Answer:

Exact Form: √13/13

Decimal Form: 0.27735009…

Step-by-step explanation:

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3 years ago

Aria drank 500 milliliters of water after her run. Her best friend, Andrea, drank 0.75 liter of water. Who drank more?

Paul [167]

To answer this item, let us convert first one of the given dimension such that both numbers will take only the same units for easier comparison. It should be noted that every liter (L) is consists of 1000 mL.

Aria: (500 mL)(1 L/1000 mL) = 0.5 L
Andrea: 0.75 L

Since, 0.75 L is higher compared to 0.5L, the answer to the item is Andrea.

8 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