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3 years ago

Help me w/ both of them pls The last one is just for me to laugh heheehe

Mathematics

2 answers:

Klio2033 [76]3 years ago

8 0

16, 87. Hope this helps

elena55 [62]3 years ago

4 0

1. 16, the ones with fruits
2. 87, with the cars

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If Dion was to purchase all 171 ticktes, then it would cost him $25.

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Answer:

He would need to purchase 8 ticket books. 8*24=192

Step-by-step explanation:

171/24=7.125 so he would need 8 books

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10x3 tens in unit and standard form

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10*3 tens= 30= 300 units

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The coordinates of the vertices of quadrilateral ABCD are A(−4, −1), B(−1, 2), C(5, 1), and D(1, −3).

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Answer:

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Step-by-step explanation:

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3 years ago

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Find the y-intercept of the line who's equation is Y = 3.5 x- 0.4.

agasfer [191]

$\sf y=mx+b$

m is the slope
b is the y-intercept

so that means that the y-intercept of

$\sf y=3.5x + .4$

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3 years ago

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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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3 years ago