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3 years ago

CAN SOMEBODY EXPLAIN THESE TO ME PLEASE? I NEED HELP

Mathematics

1 answer:

Andrei [34K]3 years ago

5 0

Answer:

Step-by-step explanation: first one is (d) you multiply base and hight and than multiply it with 12 which will convert it into inches

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The cash price of a bike is $3000. The bike can be bought on hire purchase by making a 25% down-payment and monthly payments of

horrorfan [7]

25% downpayment of $750 + $125 x 24 months

= $3750

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3 years ago

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Who is in 6th grade or 7th? <br> cause I swear hrw.com is horrible

LiRa [457]

Answer:

Not me

Step-by-step explanation:

im in high school lol

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2 years ago

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It takes exactly 26 paper clips laid end to end to make 17 7/8 inches. How long is each paper clip? (I really need the explanati

solong [7]

Answer:

11/16 inch or 0.6875 inch

Step-by-step explanation:

26 * x = 17 7/8

17 7/8 = 143/8

143/8 ÷ 26 = 143/8 x 1/26 = 143/208 = 0.6875

0.6875 in = 11/16 in

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

Check

11/16 x 26 = 17.875 or 17 7/8

4 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:

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

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3 years ago

The table shows a linear relationship between x and y.

vovangra [49]

Answer:

Step-by-step explanation:b

4 0

2 years ago